Solutions Manual for Emmy Noether’s Wonderful Theorem,
Second Edition
Dwight E. Neuenschwander
Department of Physics, Southern Nazarene University, Bethany OK 73008
(Dated: December 20, 2018)
1
Note from the author: Here I offer some detailed solutions to exercises in the second
edition of Emmy Noether’s Wonderful Theorem. I hope you don’t need to consult them too
often, but they are available for those moments when you get stuck–as I have been many
times. Your comments and corrections are always welcome.
THIS DOCUMENT IS A WORK IN PROGRESS. This version has so far solutions to
the ”Exercises” in Chapters 1-5. Over the next several weeks I aim to provide solutions
to the Exercises on Chapters 6-10. Then we’ll see about the ”Questions for Reflection and
Discussion.” Some of them have no simple or unique answers, but are intended to raise
thought-provoking issues. Your comments and corrections are welcome.
This solutions manual offers an opportunity to make some corrections and clarifying
comments about passages in the text. A couple of examples are illustrated at the end of
these introductory remarks. No matter how many times one looks at one’s own work, some
errors always sneak through! One of my professors once said to our class, as we were using
his textbook, that the only way to catch all errors is to have two people sit across a table
from one another. One follows along while the other reads aloud, character by character–
going backwards. Given the deadlines in writing and producing a book, such an effective
though exhaustive process seldom happens.
Thank you for joining me in this journey of making Emmy Noether one of our intellectual
companions.
Erratum:
* In Eq. (9.91) of Ch. 3 in the book, the right-hand side should read (ni − no )/R instead
of (no − ni )/R.
* Serious typo alert: The integral sign and dt are missing from the right-hand side of the
book’s Eq. (3.100). That equation should read:
Z t2 Z τ2 r
dxµ dxν
1 2
gµν
dτ ≈
mΦ − mv
dt.
m
dτ dτ
2
t1
τ1
2
(1)
I. WHEN FUNCTIONALS ARE EXTREMAL
CHAPTER 1: Symmetry
Exercise 1.1. Show how to express Newton’s second law, F = dp/dt, in terms of the
change in
(a) Mechanical Energy:
Multiply F = dp/dt by the displacement vector dr = vdt = (p/m)dt, using the scalar
product,
dp p
· dt
dt m
1
=
p · dp
m
F · dr =
then integrate from position a to b to obtain
2
Z b
p
F · dr = ∆
2m
a
(2)
where ∆ means final minus initial. This is the work-energy theorem; the integral of
F · dr is the work done by the net force F. For forces whose work is path-independent (i.e.,
forces whose curl vanishes everywhere), their work equals the decrease in a potential energy
function U. Letting WN C be the path-dependent work, i.e., the work done by so-called
“nonconservative” force, the work-energy theorem becomes
WN C = ∆(K + U )
(3)
where K ≡ p2 /2m denotes the kinetic energy. Letting E ≡ K + U , the total mechanical
energy, we have WN C = ∆E, implying that mechanical energy is conserved if and only if
zero work is done by nonconservative forces.
(b) Angular Momentum: Let r be a vector that locates the particle relative to some
designated point. Take the vector product of the equation of motion with r:
r×F=r×
dp
.
dt
(4)
On the right-hand side, note that
d
dp
(r × p) = v × (mv) + (r ×
)
dt
dt
3
(5)
and the first term vanishes, because v × v = 0. Therefore
r×F=
dL
dt
(6)
where r × F is the net torque, and L ≡ r × p the angular momentum, both evaluated with
respect to the designated point. Torque equal to the rate of change of angular momentum
is the rotational or orbital version of Newton’s second law.
(c) In exercise 1.1a, the connection between the conservation of energy and a symmetry
is not yet obvious for a general case. Let us try another approach by taking the derivative
of E = K + U with respect to time:
dr
∂U
dE
p dp
=
·
+ ∇U ·
+
dt
m dt
dt
∂t
∂U
dp
+ ∇U +
.
= v·
dt
∂t
By virtue of Newton’s second law, −∇U = dp/dt, the first term vanishes, so that
dE
∂U
=
.
dt
∂t
(7)
This says the mechanical energy is conserved if and only if the potential energy does not
depend explicitly on time. In that case the interaction is unchanged, or invariant, under a
time translation. For example, the non-conservative force of kinetic friction acts opposite
the direction of motion and depends on velocity. As the motion proceeds, the force changes
with velocity, so energy is not conserved due to the force changing as time elapses.
In exercise 1.1b, angular momentum about the designated point will be conserved if and
only if the net torque about that point vanishes. For L to be constant, its direction stays
fixed, and thus the particle moves in the plane defined by r and p. If a set of coordinate axes
is marked out on that plane, the conserviation of angular momentum still holds even if those
axes are rotated. Thus we see the suggestion that the conservation of angular momentum
may be related to invariance under rotations.
Exercise 1.2. A proton of mass m, at rest in the lab frame, is struck by an incoming
proton that carries incident kinetic energy K and momentum p. We are to find the minimum
K such that after the collision the system has three protons and an antiproton.
4
From special relativity, the invariance of the energy-momentum four-vector enables us
to write for the lab frame (unprimed quantities) and for the center-of-mass frame (primed
quantities)
2
2
Esys
− (psys c)2 = E 0 sys − (p0 sys c)2 ,
(8)
where “sys” refers to the system. Before the collision, the system consists of two protons;
after the collision, the system consists of three protons and one antiproton. Since energy and
momentum are separately conserved, in each frame we can use either the “before” or “after”
values of these quantities. Using the “before” values for the lab frame, and the “after” values
for the CM frame, and absorbing the c’s into the dynamical variables, we have
(2m + K)2 − p2 = (4m + K 0 )2 − 02
(9)
where K 0 denotes the kinetic energy of the three protons and one antiproton after the collision
in the CM frame. Since the total momentum in the CM frame vanishes, the minimum value
possible for K 0 is zero, which makes K its minimum value in the lab frame. Setting K 0 = 0,
making use of E 2 − p2 = m2 for the incoming proton in the lab frame, where E = K + m,
gives K = 6m (or K = 6mc2 in conventional units). it takes more tahn 2mc2 in the lab
frame to create the pp̄ pair, because momentum must be also be conserved.
5
CHAPTER 2: Functionals
Exercise 2.1. A light ray traveling at angle α above the x -axis encounters at the x = 0
plane a medium of refractive index n = no (1 + x/a) where no and a are constants. We are
to find the time required for the ray to travel from the origin to (a, y(a)) along alternative
paths. We invoke Fermat’s principle by minimizing integral
Z
Z
no a x p
1 (a,y(a))
n(x, y) ds =
1+
1 + (dy/dx)2 dx.
∆t =
c (0,0)
c 0
a
(a) With y = x,
no
∆t =
c
Z
0
a
1+
x √
3 no a
no a
2 dx = √
≈ 2.12
.
a
c
2 c
(10)
(11)
(b)
h
x
i
y = a cosh
−1
a
(12)
produces dy/dx = sinh(x/a). With the identity cosh2 (θ) − sinh2 (θ) = 1 and the change of
variable u = x/a we have
Z
no a 1
∆t =
(1 + u) cosh u du
c 0
no a
=
(2 sinh 1 − cosh 1 + 1)
c
no a
≈ 1.81
.
c
Exercise 2.2. This problem explores different meanings of the word “action” as used in
mechanics.
(a) Starting wth
Z
T
(K − U ) dt,
Γ=
(13)
0
using E = K + U gives in the integrand K − U = 2K − E, and thus
Z T
Z T
E dt.
2K dt −
Γ=
0
(14)
0
(b) Since K = 21 mv 2 = 21 mv(dx/dt)2 , we may write 2K dt = mv(dx/dt) dt = p dx and thus
Z T
Z T
Γ=
p dx =
E dt.
(15)
0
0
(c) The differential version of the last equation reads
dΓ = p dx − E dt,
6
(16)
which suggests that Γ = Γ(p, E), in which case
∂Γ
∂Γ
dp +
dE.
dΓ =
∂p
∂E
(17)
Comparing the two expressions for dΓ shows that p = ∂Γ/∂x and E = −∂Γ/∂t.
(d) If E = const., then
Z
T
p dx − ET
Γ =
0
Z
T
p dx + const.
=
0
Therefore requiring Γ to be a minimum means the “abbreviated action”
RT
0
p dx must be
minimized.
Exercise 2.3. A particle of mass m falls without air resistance in a uniform gravitational
field g = g(−ĵ). We are to evaluate the functional for proposed trajectories y = y(t). We
start with the functional of Hamilton’s principle,
Z t0 1 2
Γ=
mẏ − mgy dt
2
0
(18)
where we take up to be positive. Let the particle be released from rest at the origin at
t = 0. Since it falls downward into negative y, from elementary mechanics we know that
y = − 12 gt2 , which will facilitate some discussion. (a) If y(t) = A(t/to )2 , then
Z to 2
t2
2 t
Γa =
2A m 4 − mgA 2 dt
to
to
0
mA 2A
− gto .
=
3
to
(b) With y = A(1 − e−t/to ) we find
Z to 1 A2 −2/to
−t/to
Γb =
m (e
) − mgA(1 − e
) dt
2 t2o
0
A 1 − e−2
−1
= mA
− gto e
.
to
4
Exercise 2.4 The given functional, an application of Hamilton’s principle applied to a
particle moving with potential energy 21 mω 2 x2 , describes a particle oscillating with period
T and angular frequency ω = 2π/T . Define Γ for one period:
Z T
1
1
2
2 2
Γ=
mẋ − mω x
dt.
2
2
0
7
(19)
We pretend we do not know the correct x(t) for a simple harmonic oscillator, and try different
candidate functions x(t) and see which gives the minimum value of Γ.
(a) If we assume dx/dt = +vo for the first half of a cycle, when x goes from x = −A to
x = +A (thus vo = 4A/T ), and assume dx/dt = −vo for the second half of the cycle, when
x goes from x = +A to x = −A, then the x(t) that satisfies these boundary conditions is
x = vo t − A for the first half, and x = 3A − vo t for the second half of the cycle. With this
ansatz and letting K ≡ 21 mvo2 , the functional becomes
Z T 1
1
2
2
2
2
K − mω (vo t − A)
K − mω (3A − vo t)
Γ =
dt +
dt
2
2
0
T /2
π2
= KT 1 −
12
≈ 0.1774 KT.
Z
T /2
(b) With the ansatz x = A cos(ωt) so that ẋ = −Aω sin(ωt), the functional becomes
Z T
2
1
2 2
Γ = mω A
sin (ωt) − cos2 (ωt) dt
2
0
Z T
1
2 a
cos(2ωt) dt
= − mω A
2
o
= 0.
Exercise 2.5 In this problem we consider candidate investment strategies in the hypothetical earning functional E :
Z
E=
T
[YB (t)NB (t) + YS (t)NS (t)] dt.
0
In all of the following scanarios we are to assume NB = NS = const. ≡ No /2.
(a) With YB = YS = const. ≡ Yo the integral gives E = Yo No T ≡ Eo .
(b)With YB = 2Yo and YS = Yo e−t/T , we obtain
Z
Yo No T
E =
2 + e−t/T dt
2
o −1 2−e
= Eo
2
≈ 0.816 Eo .
(c) With YB = 2Yo and YS = Yo [1 + cos(ωt)], we obtain E = Eo /2.
8
(20)
Chapter 3: Extremals
Exercise 3.1 This problem revisits the example of Eq. (3.41), about a ligh ray passing
through a medium of uniform refractive index. We are to push this problem further using
the Hamiltonian. The given Lagrangian is
L=
np
1 + y 02 = L(y 0 )
c
(21)
where y 0 = dy/dx and n is spatially uniform. (a) To find the Hamiltonian H, we first need
the canonical momentum p = ∂L/∂y 0 , which yields
y0
p=λ p
1 + y 02
(22)
where λ ≡ n/c. Therefore we write H = y 0 p − L in terms of p, which gives
p
H = − λ2 − p2 .
(23)
(b) Next we ask whether the form of the Euler-Lagrange equation
∂L
dH
=−
dx
∂x
(24)
yields a straight line. We find
1
dH
=−
dx
c
dn
dx
p
1 + y 02
(25)
which vanishes since n is spatially uniform. Therefore H = const. and thus p2 = const.,
which in turn means y 0 = const. and this integrates to the equation of a straight line.
Exercise 3.2 This problem uses Fermat’s principle to derive the “lens maker’s equation.”
Refer to Fig. 3.4 on page 59. [Note that for the figure and Eq. (3.91) to be consistent, the
distances “o” and “i ” shown in the figure should be denoted so and si respectively.] Let lo
denote the distance from the object to point A, and let li denote the distance from A to the
image. By Fermat’s principle, we need to make an extremum the quantity
c∆t = no lo + ni li
(26)
where
lo =
p
(so + R)2 + R2 − 2R(so + R) cos ϕ
9
(27)
and
li =
p
(si − R)2 + R2 − 2R(si − R) cos ϕ.
(28)
Setting d(∆t)/dt = 0 gives
no
ni
(so + R) = (si − R).
lo
li
(29)
For small ϕ, lo ≈ so and li ≈ si , so that
no ni
ni − no
+
≈
.
so
si
R
(30)
Note: Eq. (9.91) contains a typographical error. The right-hand side should read (ni −no )/R
instead of (no − ni )/R.
Exercise 3.3 This problem asks us to find the trajectory of a light ray passing through
a refractive medium that exhibits a gradient in the vertical direction. Our functional is
Z t
y p
nw + no (1 − )
c∆t =
1 + y 02 dx
(31)
h
0
where y 0 ≡ dy/dx. The Euler-Lagrange equation,
d ∂L
∂L
=
∂y
dx ∂y 0
(32)
becomes
−
where s ≡
no s
no y 02
n2 y 02
=−
+ ny 00 − 2
h
hs
s
(33)
p
1 + y 02 . This nonlinear differential equation is a bit cumbersome, so let’s try
the other form of the Euler-Lagrange equation,
∂L
dH
=−
∂x
dx
(34)
where H is the Hamiltonian, H = y 0 (∂L/∂y 0 ) − L = −n/s. This second form of the ELE
becomes
1 dn
1 ds
=
n dx
s dx
(35)
which integrates to n = κs where κ = const. To determine κ, invoke the boundary condition
y 0 = 0 when x = 0. Then
κ = nw + no (1 − α).
With κ known we return to n = κs, which may be written
r dy
n 2
=±
− 1.
dx
κ
10
(36)
(37)
With the change of variable u = n/κ, and assuming u2 ≥ 1, we obtain the integral
Z
Z
no
du
√
=±
dx
κh
u2 − 1
(38)
which integrates to
cosh−1 u = ±
no x
+β
κh
(39)
where β = const. The boundary condition y = αh when x = 0 gives β = 0. Thus
n x
o
u = cosh
κh
(40)
and finally
y=
n x i
h h
o
nw + no − κ cosh
.
no
κh
(41)
Exercise 3.4 (a) To measure the shortest distance between two points on a curved surface
(or, more generally, the “geodesic” between two points in a “curved space”) begin with the
√
distance Lagrangian L = gµν uµ uν , where uµ = dxµ /dσ ≡ ẋµ . The coefficients gµν are
the components of the metric tensor. In 3-D Euclidean space L is the familiar Pythogorean
theorem,
L = [ẋ2 + ẏ 2 + ż 2 ]1/2 .
(42)
Since ∂L/∂xµ = 0, it follows from the ELE that ∂L/∂ ẋµ = ẋµ /L = const. for µ = 1, 2, 3.
Write ẋ/L = a, ẏ/L = b, and ż/L = c where a, b, c are constants. From this we have
ẋ = La = ẏ
a
b
(43)
which integrates to x = (a/b)y + const., and similarly for x and z, and for y and z. Thus
the equation of the geodesic in 3-D Euldidean space is a straight line.
(b) In cylindrical coordinates (r, θ, z) the Lagrangian is
L = [ṙ2 + r2 θ̇2 + ż 2 ]1/2 .
For motion on a cylinder of radius R, the Lagrangian reduces to L =
(44)
p
R2 θ̇2 + ż 2 . Since
∂L/∂θ = 0 and ∂L/∂z = 0, it follows that R2 θ̇/L = ` and ż/L = ζ where ` and ζ are
constants. Therefore θ̇/ż = const. which gives ∆θ = const. × ∆z, the equation of a helix
on the cylinder’s surface.
11
(c) For the Schwarzschild geometry around a star of mass M, the metric, in spherical spatial coordinates, where t denotes the time coordinate, the proper time interval dτ is
given by
dτ 2 = Adt2 +
dr2
− r2 dθ2 − (r2 sin2 θ)dϕ2
A
(45)
where
A = A(r) = 1 −
rs
r
(46)
and where rs ≡ 2GM/c2 is the “Schwarzschild radius” of the mass M, with G denoting
Newton’s gravitational constant and c the speed of light in vacuum. The functional may be
written
Z
Γ=
a
b
Z br
dxµ dxν
dτ
dτ =
gµν
dτ dτ
a
(47)
with Lagrangian
ṙ2
L = Aṫ −
− r2 θ̇2 − (r2 sin2 θ)ϕ̇2
A
2
1/2
(48)
where ẋµ ≡ dxµ /dτ , the derivative with respect to proper time. Comparing these two
expressions for Γ shows that, numerically, L = 1 even though it is a function of r, θ, ṫ, ṙ, θ̇,
and ϕ̇.
The trajectory xµ = xµ (τ ) of a freely-falling particle of mass m will be the solution of
the Euler-Lagrange equation,
∂L
d ∂L
=
.
µ
∂x
dτ ∂ ẋµ
(49)
Let’s begin by considering canonical momenta that are constant. Noting that ∂L/∂t = 0,
we have
∂L
E
= Aṫ ≡ .
(50)
m
∂ ṫ
We call this constant E/m, the energy per mass of m, because in the limit of special relativity,
const. =
A = 1 and dτ = (1 − v 2 /c2 )− 1/2 ≡ γSR = E/mc2 in conventional units. Now in the
Schwarzschild metric, dt/dτ = A−1/2 (1 − v 2 /Ac2 )−1/2 , and the falling “test particle” of mass
m interacts with the gravitational source mass M. Therefore in the low-speed, weak-field
limit, mAṫ reduces to, in conventional units, to
√
2
dt
mc
A
1
GM m
=q
≈ mc2 + mv 2 −
mc2 A
dτ
2
r
v2
1 − Ac
2
(51)
to first order in Newtonian kinetic and potential energies. Let the particle begin its motion
at r → ∞, in the flat spacetime of special relativity for which A = 1, dt ≈ dτ , and E = mc2 .
12
Then (with c = 1)
Aṫ = 1.
(52)
Next, we note that ∂L/∂ϕ = 0, and thus
const. =
`
∂L
= r2 sin2 θ ϕ̇ ≡
∂ ϕ̇
m
(53)
where we have made use of the fact that, numerically, L = 1. We call this constant `/m the
angular momentum per unit mass, because in central force motion of classical mechanics,
the angular momentum ` = mr2 ϕ̇ is conserved, and thus the particle moves in a plane, which
can be taken to be the xy plane. In our case we note that a ≡ r sin θ is the distance of the
particle from the z -axis, so that here we have ` = ma2 ϕ̇. Similar to the classical mechanics
case, let the particle’s trajectory lie in the xy plane, where θ = π/2. Now the Lagrangian of
Eq. (48) may be written
1/2
ṙ2
2 2
− r ϕ̇
(54)
L = Aṫ −
A
and the rest of the story of the trajectory may be found by solving for r(τ ) from the Euler
2
Lagrange equation,
∂L
d ∂L
=
.
(55)
∂r
dτ ∂ ṙ
Using Aṫ = 1, ϕ̇ = `/mr2 , and noting that dA/dr ≡ A0 = rs /r2 and dA/dτ ≡ Ȧ = ṙA0 , the
Euler-Lagrange equation for r gives, after some algebra,
r̈ +
` A
1 A0
(1 − ṙ2 ) − 2 3 = 0.
2A
m r
(56)
For simplicity, consider the special case of the particle of mass m released from rest at
infinity, and allowed to free-fall radially towards the star. Then ` = 0 and the EulerLagrange equation for r reduces to
r̈ = −
1 A0
(1 − ṙ2 ).
2A
(57)
Writing r̈ = dṙ/dτ = ṙdṙ/dr, this separates into
ṙdṙ
1 dA
−−
.
2
1 − ṙ
2 A
(58)
With the change of variable u = 1 − ṙ2 this easily integrates to ln u = ln A + ln C where C
is an integration constant. The initial conditions at infinity require C = 1, and therefore
1 − ṙ2 = A, or
r
ṙ = −
13
rs
r
(59)
where the minus sign is chosen for the square root because the particle moves from large to
small values of r. Notice that in the freely-falling observer’s frame, the event horizon (where
r = rs ) rushes towards that observer at speeds approaching that of light. Such is not the case,
however, for another observer observing from afar, where (t, r, θ, ϕ) coordinates, collected
locally for each event, are analyzed later to reconstruct the events. Since dt/dτ = 1/A,
dr/dτ can be re-written (dr/dt)ṫ = (dr/dt)A−1 to give
r
r
rs
rs rs
dr
= −A
=− 1−
.
dt
r
r
r
(60)
Thus to this remote observer, the velocity of an infalling particle approaches zero as the
particle’s r -coordinate approaches the event horizon.
Exercise 3.5 In a system of Euclidean coordinates X λ we are given
d dX λ
=0
dσ
dσ
(61)
and are to show how it transforms to a more general curvilinear system of coordinates xµ ,
where X λ = X λ (xµ ). By the chain rule,
d ∂X λ dxµ
0 =
dσ ∂xµ dσ
∂ 2 X λ dxν dxµ ∂X λ d2 xµ
=
+
.
∂xν ∂xµ dσ dσ
∂xµ dσ 2
Multiply by ∂xρ /∂X λ :
∂xρ ∂ 2 X λ
∂X λ ∂xν ∂xµ
dxν dxµ
+
dσ dσ
∂xρ ∂X λ
∂X λ ∂xµ
d2 xµ
= 0.
dσ 2
(62)
Recognizing
∂xρ ∂X λ
= δρµ
∂X λ ∂xµ
(63)
yields Eq. (3.95).
Exercise 3.6 (a) We are to find the Lagrangian whose Euler-Lagrange equation yields
the equation of motion of the damped harmonic oscillator,
−kx − bẋ = mẍ.
14
(64)
Clearly if b = 0 the equation reduces to that of a simple harmonic oscillator, whose Lagrangian is
1
1
Lo = mẋ2 − kx2 .
2
2
(65)
Let us modify this by rescaling Lo with a function f = f (t, x, ẋ) so that L = Lo f , and
determine the unknown function f by fitting the given equation of motion to the EulerLagrange equation,
∂L
d
−
∂x dt
∂L
∂ ẋ
.
(66)
Inserting L = Lo f into the Euler-Lagrange equation gives
∂Lo
∂f
d
∂Lo
∂f
+ Lo
−
f + Lo
= 0,
∂x
∂x
dt
∂ ẋ
∂ ẋ
(67)
which may be rearranged into
f˙
−kx − mẍ − mẋ
f
!
f + Lo
d ∂f
∂f
−
∂x dt ∂ ẋ
= 0.
(68)
This will reduce to the desired equation of motion if ∂f /∂x = 0, and ∂f /∂ ẋ = 0, and
−b = mf˙/f which integrates to f (t) = ebt/m . Thus the desired Lagrangian is
1
1 2 bt/m
2
L(t, x, ẋ) =
mẋ − kx e
.
(69)
2
2
(b) The canonical momentum is
p=
∂L
= mẋebt/m .
∂ ẋ
(70)
so ẋ = (p/m)e−bt/m . Therefore,
H = ẋp − L
2
p2 −bt/m
p −2bt/m 1 2 bt/m
=
e
−
e
− kx e
m
2m
2
2
p −bt/m 1 2 bt/m
=
e
+ kx e
.
2m
2
(c) Since ∂L/∂x 6= 0, the canonical momentum is not conserved. Since ∂L/∂t 6= 0, the
Hamiltonian is not conserved. Even though momentum and mechanical energy are not
conserved, after developing Noether’s theorem we will find a linear combination of p and H
that is conserved.
15
Exercise 3.7 Here we are confronted with a two identical physical pendulums in tandem,
one attached to a pivot point and the other attached to the bottom of the first pendulum.
They swing in a plane. We are to find the equations of motion through the Lagrangian
formalism. Let I be the moment of inertia of one pendulum about the suspended end, so
that
Z
I=
r2 dm
(71)
where r denotes the distance of the mass element dm from the suspended end. To form the
Lagrangian, we need the kinetic energy and potential energy of the double pendulum. Let
θ1 and θ2 be the angles from the vertical made by the top and lower pendulum respectively.
Set point A be the top suspension point, and B be the point where the two pendulum are
joined together.
For the top pendulum’s kinetic energy we may write K1 = 21 I θ̇12 . The bottom pendulum’s
kinetic energy does not end with K2 = 21 I θ̇2 2 –it could have kinetic energy even if θ̇2 = 0
because the top pendulum carries the bottom one along. We can calculate K2 as follows:
Z
1
(dm)v22 (r)
K2 =
(72)
2
where (according to Newtonian relativity)
v2 (r) = v(dm
relative to B)
+ v(B
relative to A)
= r2 θ̇2 θ̂ 2 + `θ̇1 θ̂ 1
where r2 is the distance of dm from B. Therefore
v22 = r2 2 θ̇22 + `2 θ̇1 2 + 2`r2 θ̇1 θ̇2 (θ̇ˆ1 · θ̇ˆ2 ).
(73)
ˆ ˆ
Noting that θ̇ 1 · θ̇ 2 = cos(θ2 − θ1 ), we obtain
Z
Z
Z `
Z `
1 2 2 `
1 2 ` 2
K2 = θ̇2
r2 dm + ` θ̇1
dm + `θ̇1 θ̇2
r2 dm + 2`θ̇1 θ̇2 cos(θ2 − θ2 )
r2 dm. (74)
2
2
0
0
0
0
The first integral gives lower pendulum’s moment of inertia I about its pivot point; the
second integral gives the pendulum’s mass m, and the third integral can be written in terms
of the pendulum’s center of mass, located at the distance ro from the pivoted end,
Z `
r2 dm = mro .
0
16
(75)
Therefore
1
1
K1 + K2 = (I + m`2 )θ̇12 + I θ̇22 + 2m`ro θ̇1 θ̇2 cos(θ2 − θ1 ).
2
2
(76)
Turning to the potential energy U = U1 +U2 , let us take the zero of gravitational potential
energy to be the point at distance ` below the upper pivot point, i.e., point B when θ1 = 0
(the y-axis). For a uniform gravitational field of magnitude g, we have for the potential
energy of the upper rod,
dU1 = (dm)gy1
= (dm)g(` − r1 cos θ1 )
which integrates to
U1 = mg(` − ro cos θ1 ).
(77)
As a consistency check, note that at θ1 = 0, U1 = mg(` − ro ) which would equal 21 mg` for
the special case of a thin uniform rod.
Turning to U2 = (dm)gy2 where
y2 = `(1 − cos θ1 ) − r2 cos θ2 ,
(78)
we find that U2 = mg`(1 − cos θ1 ) − mgro cos θ2 . Thus U = U1 + U2 becomes
U = mg`(2 − cos θ1 ) − mgro (cos θ1 + cos θ2 ).
(79)
As a consistency check, a pair of thin uniform rods that this becomes
U = −mg(−2` +
3
1
cos θ1 + cos θ2 ),
2
2
(80)
so when θ1 = θ2 = 0 then U = 0, as expected; and when θ1 = θ2 = 90o then U = 2mg`, also
as expected.
Finally we have the Lagrangian L = K − U for the dual-pendulum system. Letting
I 0 ≡ I + m`2 ,
1
1
L(θ1 , θ2 , θ̇1 , θ̇2 ) = I 0 θ̇12 + I θ̇22 − mg[2` − cos θ1 ) + mgro (cos θ1 + cos θ2 ).
2
2
(81)
The Euler-Lagrange equation for θ1 ,
∂L
d ∂L
=
∂θ1
dt ∂ θ̇1
17
(82)
and similarly for θ2 gives the coupled equations
m`ro cos δ
mg(` + ro )
sin θ1 = −
θ̈2
θ̈1 +
I0
I0
(83)
and
θ̈2 +
mgr o
I
sin θ2 = −
m`ro cos δ
I
θ̈1
(84)
where δ ≡ θ2 − θ1 .
Exercise 3.8 A plane pendulum of variable length is realized in a linear spring of negligible mass and constant k, from which a particle of mass m is suspended.
(a) To write the Lagrangian, map the system with plane polar coordinates (r, θ):
1
1
1
L(r, θ, ṙ, θ̇) = mṙ2 + mr2 θ̇2 − k(r − ro )2 − mgr(1 − cos θ)
2
2
2
(85)
where ro denotes the relaxed length of the spring. The Euler-Lagrange equation for r gives
r̈ + ωo2 (r − ro ) + g(1 − cos θ) = rθ̇2
where ωo ≡
(86)
p
k/m. As a consistency check, if θ = 0, letting ≡ r − ro turns this into
¨ + ωo2 = 0,
(87)
the expected equation of harmonic motion for a mass on a spring.
Turning to the Euler-Lagrange equation for θ, it becomes
θ̈ +
g
2ṙθ̇
sin θ = −
.
r
r
(88)
If r = const. this becomes the expected equation for the plane pendulum,
θ̈ +
g
sin θ = 0.
r
(89)
(b) When the pendulum’s motion is not confined to a plane, let the suspension point be
the origin of a system of spherical coordinates (r, θ, ϕ). The pendulum hangs below the xy
plane, so will be convenient to define ψ ≡ π − θ. In spherical coordinates, the distance ds
between two nearby points follows from
ds2 = dr2 + r2 dφ2 + r2 dϕ2 sin2 ψ
18
(90)
and thus the velocity squared of the hanging point mass m is
v 2 = ṙ2 + r2 ψ̇ 2 + r2 ϕ̇2 sin2 ψ.
(91)
Taking the zero of gravitational potential energy to be the xy plane with positive potential
energy on the +z side, the total potential energy is
1
U = k(r − ro )2 − mgr cos ψ.
2
(92)
1
1 L(r, ṙ, ψ, ψ̇, ϕ) = m ṙ2 + r2 ψ̇ 2 + r2 ϕ̇2 sin2 ψ − k(r − ro )2 + mgr cos ψ.
2
2
(93)
Therefore the Lagrangian is
Because ∂L/∂ϕ = 0, the Euler-Lagrange equation for ϕ yields a conservation law,
mr2 ϕ̇ sin2 ψ = const. Since r sin ψ is the perpendicular distance from the z -axis to mass
m, this constant is the z -component of the particle’s angular momentum `z .
The Euler-Lagrange equation for ψ,
d ∂L
∂L
=
∂ψ
dt ∂ ψ̇
(94)
r sin ψ(rϕ̇2 cos ψ − g) = 2rṙψ̇ + r2 ψ̈.
(95)
becomes
or
cos θ `2z
− rg sin θ = 2rṙψ̇ + r2 ψ̈.
(96)
2
2
2
sin θ m r
As a consistency check if ṙ = 0 and ϕ̇ = 0 this reduces to the usual equation for a simple
plane pendulum of constant length.
The Euler-Lagrange equation for r yields
`2z
2
r ψ̇ +
− ωo2 (r − ro ) + g cos ψ = r̈,
mr3 sin2 θ
(97)
(where ωo2 ≡ k/m) which reduces to the expected equation of motion for a mass on a vertical
spring, suspended in a uniform gravitational field, when ψ = 0 = const. and ϕ̇ = 0.
Exercise 3.9 Dido’s problem: Find the function y(x) in the xy plane that encloses the
maximum area with a fixed perimeter S. The perimeter forms some closed path that crosses
the x -axis at x = a and at x = b. The area A is
Z b
A=
y(x) dx
a
19
(98)
subject to the constraint of a fixed perimeter, which is
I p
S=
dx2 + dy 2 dx.
or
Z b p
1 + y 02 −
a
S
b−a
(99)
dx
where y 0 = dy/dx. Thus we form the constrained functional
p
Z b
S
02
Γ=
y−λ
1+y −
dx
b−a
a
with λ the Lagrange multiplier. The Euler-Lagrange equation becomes
"
#
d
−λy 0
p
1=
.
dx
1 + y 02
(100)
(101)
(102)
Upon integrating and doing some algebra, we find
√
±x =
λ2 − xx
dy
.
dx
(103)
Another integration produces
x2 + y 2 = λ 2 ,
(104)
the equation of a circle of radius λ.
Exercise 3.10 We are to find the shape of a flexible uniform cord that hangs suspended
from two fixed points of support in a uniform gravitational field g (the curve is called a
catenary if the support points are at the same elevation). Since there is no kinetic energy, by
Hamilton’s principle only the potential energy remains to be minimized. Let the cord hang
in the xy plane, with y vertical, with the suspension points above x = 0 and x = a. A string
segment of mass dm contributes an increment of the string’s potential energy, dU = (dm)gy
p
p
where dm = µds, with µ the mass per unit length, and ds = dx2 + dy 2 = 1 + y 02 dx
where y 0 ≡ dy/dx. So the potential energy is
Z a p
U =µ
y 1 + y 02 dx,
(105)
0
setting up the Lagrangian L(y, y 0 ) = y
p
1 + y 02 . The Ḣ version of the Euler-Lagrange
equation is convenient here, where it takes the form
−
∂L
dH
=
∂x
dx
20
(106)
yielding H = const. ≡ C. By definition, H ≡ y 0 (∂L/∂y 0 ) − L, which after some algebra
yields
dy
=
dx
r
y2
− 1.
C2
(107)
Let u ≡ y/C. Now the differential equation can be integrated via separation of variables:
Z
Z
du
1
√
=
dx
(108)
C
u2 − 1
which integrates to
C cosh−1 u = x + const.
(109)
or
y = C cosh
x
C
+κ
(110)
follows at once.
Exercise 3.11 This is the celebrated brachistochrone problem: find the shape of the
path along which a particle could slide without friction in a uniform gravitational field, to
make the trip between the two fixed points in the least time. Rearrange the definition of
speed, v = ds/dt where ds denotes an arc length, into
p
1 + y 02
dt =
dx,
v
(111)
where y 0 ≡ dy/dx. Orient the coordinate axes with x vertically down in the direction of the
gravitational field, so y is horizontal. By the conservation of energy E,
1
E = mv 2 + mgx.
2
(112)
Let the desired curve pass through the origin, from where the particle is released from rest,
√
so that E = 0 and v = 2gx. The time for a slide from x = 0 to x = a will be
Z ap
1 + y 02
√
∆t =
dx,
(113)
2gx
0
suggesting the Lagrangian
r
0
L(x, y ) =
1 + y 02
.
x
(114)
Since ∂L/∂y = 0, it follows that ∂L/∂y 0 = const., which produces as a preliminary step
y0
1
p
=√
02
κ
x(1 + y )
21
(115)
where κ is some constant. The variables can be separated into the integral
Z √
x dx
√
.
y=
κ−x
(116)
More than one option presents itself. If we integrate this using, say, an integral table, we
obtain
√
y = − κx −
x2
κ
− sin−1
2
κ − 2x
κ
+
κπ
4
(117)
which is fine, but does not make our path’s shape easy to visualize! Here is another approach
(as seen in most textbooks): Return to the integal over x and multiply numerator and
√
denominator of the integrand by x, and let κ ≡ 2a, which gives the integral the form
Z
x dx
√
y=
.
(118)
2ax − x2
Now let x = a(1 − cos θ). The integral nicely simplifies to
Z
y = a (1 − cos θ) dθ
(119)
so that y = a(θ − sin θ). The two parametric equations for x and y in terms of θ describe a
cycloid, the trajectory swept out by a point on a circle of radius a that rolls without slipping.
θ is the angle through which the circle has rolled from its starting-place. We take as initial
condition θ = 0 when the point coincides with the origin.
Exercise 3.12 It is claimed in special and general relativity that the path of a particle
Rb
falling freely in a graviational field is given by a dτ is a maximum. We seek an argument
to justify this principle.
(a) In special relativity, dτ 2 = dt2 − dx2 so that
Z
Γ=
b
(dt2 − ds2 )1/2
(120)
a
where ds denotes the spatial distance between events a and b in the reference frame where
the time between them is dt (in units where time is measured in meters). dτ is an invariant
between all inertial reference frames in special relativity, and numerically equals the proper
time between events 1 and 2, because in that frame ds = 0. With this in mind, watch a
billiard ball at rest on the billiard table. Event 1 is when you start watching, at time t = 0.
Event 2 is when you stop watching, at time t = T . The ball sits at the same location in
22
your reference frame, at both events 1 and 2. The proper time between these two events is
therefore ∆τ 2 = T 2 . If the ball moved through spacetime along any other path, then in your
reference frame that path would also involve motion through space, with a spatial interval
∆s which would make ∆τ 2 = T 2 − ∆s2 < T 2 . If ∆τ were required to be a minimum, that
would not rule out ∆t2 − ∆s2 = 0, suggesting that massive bodies could move at the speed
of light!
As for the principle applying to general relativity, we note that general relativity must
Rb
constain special relativity as a special case. A sober choice would be to postulate that a dτ
is also a maximum for a freely-falling particle in the presence of gravitation.
More generally, when the two events are widely separated in curved spacetime, one says
that the geodesic is an “extremum” rather than a “maximum.” As Misner, Thorne, and
Wheeler remark (see Gravitation p. 318), “When several distinct geodesics connect two
events, the typical one is not a local maximum (‘mountain peak’) but a saddle point (’mountain pass’)...”
(b) Serious typo alert: The integral sign and dt are missing from the right-hand side of
eq. (3.100). That equation should read:
Z t2 Z τ2 r
dxµ dxν
1 2
gµν
dτ ≈
mΦ − mv
dt.
m
dτ dτ
2
t1
τ1
(121)
Sorry about that!
In part (b) we are to justify the (corrected) equation just mentioned. Let us begin by
identifying the time and space components in the line element. Let uµ ≡ dxµ /dτ ,
gµν
dxµ dxν
= g00 (u0 )2 + gkk uk uk
dτ dτ
(122)
with sum over k = 1, 2, 3. For low speeds, dτ ≈ dt, and thus u0 ≈ 1 and uk ≈ dxk /dt = v k .
Taking as given the low-velocity, weak-field result g00 ≈ 1 + 2Φ, (This low-velocity, weakfield result for g00 is derived in most any general relativity textbook.) where Φ denotes the
Newtonian gravitational potential, we have
Z τ2 r
Z t2
1/2
dxµ dxν
m
dτ ≈ m
1 + 2Φ − v 2
dt.
gµν
dτ dτ
τ1
t1
performing a binomial expansion on the square root gives
Z τ2 r
Z t2 dxµ dxν
1 2
m
gµν
dτ ≈
m + mΦ − mv dt.
dτ dτ
2
τ1
t1
23
(123)
(124)
Since the additive m in the integrand contributes only an additive constant to the functional,
which does not affect the Euler-Lagrange equation, the functional for a particle of mass m
falling freely through a gravitational potential energy U ≡ mΦ is therefore
Z t2
1
(U − mv 2 ) dt,
Γ≈
2
t1
(125)
which has the opposite sign of the Lagrangian in Hamilton’s principle.
3.13 We are to show that the Lagrangian
L(x, ψ, ψ ∗ , ψx , ψx∗ ) = −
where ψx ≡ ∂ψ/∂x and
∗
h̄2 ∗
ψ ψx − ψ ∗ [E − U (x)]ψ
2m x
(126)
denotes complex conjugate, yields the time-independent
Schrödinger equation. The Euler-Lagrange equation for ψ,
d ∂ψ
∂L
=
∂ψ
dx ∂ψx
becomes
d
(E − U )ψ =
dx
h̄2
ψx
−
2m
(127)
(128)
which is the time-independent Schrödinger equation,
h̄2
ψxx + U ψ = Eψ.
−
2m
(129)
. The ELE for ψ ∗ gives the complex conjugate of the Schrödinger equation.
3.14 We are to verify that the Lagrangian
√
L = −m 1 − v 2 ≡ −mγ −1 ,
(130)
in units where c = 1 and where γ ≡ (1 − v 2 )−1/2 , yields the relativistic Newton’s second law
for a free particle. Notice how this Lagrangian comes from the functional
Z
Γ = −m dτ
Z √
= −m
dt2 − ds2
Z √
= −m
1 − v 2 dt
24
where ds denotes an increment of spatial distance and v = ds/dt. To verify (a) that the
free-particle equation of motion follows from this Lagrangian, and (b) to find the canonical
momentum, we turn to the ELE, which says
∂L
d ∂L
=
.
∂xµ
dτ ∂v µ
(131)
Since ∂L/∂xµ = 0, the canonical momentum pµ = const., where
pµ ≡
∂L
= mvµ γ,
∂v µ
(132)
the answer to part (b). For part (a), since pµ = mvµ γ = muµ = const., it follows that
duµ /dτ = 0, the relativistic generalization of Newton’s second law for free particle motion.
(c) The Hamiltonian is
H = v µ pµ − L
= mv 2 γ + mγ −1
= mγ
which will be recognized as the free particle’s total energy according to special relativity,
E = mγ (or E = mc2 γ in conventional units).
3.15 In this problem we are given the Lagrangian of a relativistic particle of mass m and
electric charge q moving through a reference frame where the electromagnetic potentials are
A and V, where in units with c = 1,
L = −mγ −1 + qA · v − qV
= −m(1 − vj v j )−1/2 + qAj v j − qV
(a) The canonical momentum pk is
∂L
∂v k
= mvk γ + qAk
pk =
so that
vk =
pk − qAk
.
mγ
25
(133)
(b) Thus the Hamiltonian H is
H ≡ v k pk − L
p − qA
p − qA
−1
=
· p + mγ − qA ·
+ qV
mγ
mγ
(p − qA)2
−1
= γ
+ m + qV.
m
(c) Since (p − qA)/mγ = v, the Hamiltonian of part (b) is numerically equal to the relativistic energy E, because
(mvγ)2
H = γ
+ m + qV
m
= mγ + qV
−1
where γ ≡ (1 − v 2 )−1/2 has been used in going from the first to the second line. mγ + qV
represents the particle’s mass and kinetic energy (mγ = m + K) plus the interaction energy
qV . Therefore
E 2 − p2 = (mγ + qV )2 − p2 .
(134)
Using p = mvγ + qA and the expression for γ in terms of the velocity, after some algebra
this becomes
E 2 − p2 = m2 + 2mγq(V − v · A) + q 2 (V 2 − A2 ).
(135)
This differs from the free-particle result E 2 − p2 = m2 with the interaction terms.
3.16 This problem asks us to generalize the Euler-Lagrange equation to Lagrangians that
include a second derivative. Consider functionals of the form
Z b
Γ=
L(t, q µ , q̇ µ , q̈ µ ) dt
(136)
a
under the variation
q 0µ = q µ + εη µ ,
(137)
where η µ = η µ (t) with η µ (t) and its derivatives vanishing on the endpoints. Thus we evaluate
δΓ, so that
dΓ
dε
Z b
0
∂L dq 0µ
∂L dq̇ 0µ
∂L q̈ 0µ
=
+
+
∂q 0µ dε
∂ q̇ 0µ dε
∂ q̈ 0µ dε
a
Z b
∂L µ ∂L µ ∂L µ
=
η + µ η̇ + µ η̈
dt.
∂q µ
∂ q̇
∂ q̈
a
26
dt
0
An integration by parts,
d
dt
∂L µ
η̇
∂ q̈ µ
∂L
d
= µ η̈ µ + η̇ µ
∂ q̈
dt
∂L
∂ q̈ µ
,
(138)
turns δΓ into
Z b
δΓ =
a
∂L µ ∂L µ
d ∂L
η + µ η̇ − η̇ µ
µ
∂q
∂ q̇
dt ∂ q̈ µ
∂L µ
dt +
η̇
∂ q̈ µ
b
.
(139)
a
The integrated boundary terms vanish, leaving
Z b
∂L µ
µ
δΓ =
dt
η + Qµ η̇
∂q µ
a
(140)
where
Qµ ≡
∂L
d ∂L
−
.
∂ q̇ µ dt ∂ q̈ µ
(141)
Now consider another integration by parts, using
d
(Qµ η µ ) = Qµ η̇ µ + η µ Q̇µ .
dt
so that
Z b
δΓ =
a
∂L
− Q̇µ
∂q µ
dt + Qµ η µ |ba .
(142)
(143)
The boundary terms again vanish, so δΓ = 0 if and only if the remaining integrand vanishes:
dQµ
∂L
−
µ
∂q
dt
∂L
d ∂L
d ∂L
=
−
−
,
∂q µ dt ∂ q̇ µ dt ∂ q̈ µ
0 =
which is the third-order Euler-Lagrange equation, for Lagrangians that include q̈ µ :
d ∂L
d2 ∂L
∂L
−
+
= 0.
∂q µ dt ∂ q̇ µ dt2 ∂ q̈ µ
(144)
(b) Now we are to generalize the Euler-Lagrange equation to Lagrangians of the form
...
L = L(q, q̇, q̈, q , . . . , q (n) )
(145)
where
q
(n)
dn q
≡ n.
dt
(146)
In what follows it proves convenient to drop the coordinate subscripts and superscripts. To
further simplify the notation, let
Ln ≡
27
∂L
∂q (n)
(147)
with
∂L
.
(148)
∂q
Under the variation q 0 = q + εη where η and its derivatives vanish on the boundaries, recall
L0 ≡
that in the n = 3 encountered above, we saw that
Z b
...
(L0 η + L1 η̇ + L2 η̈ + L3 η ) dt.
δΓ =
(149)
a
Using the product rule for derivatives on d(L3 η)/dt and setting the integrated boundary
terms to zero, δΓ becomes
δΓ =
Z b
L0 η + L1 η̇ + [L2 − L̇3 ]η̈
dt.
(150)
a
Let
so that
Z
Q2 ≡ L2 − L̇3
(151)
(L0 η + L1 η̇ + Q2 η̈) dt.
(152)
b
δΓ =
a
With another integration by parts on d(Q2 η̇)/dt, and again noting that the boundary terms
vanish, we obtain
δΓ =
Z b
h
i L0 η + L1 − Q̇2 η̇ dt.
(153)
Q1 ≡ L1 − Q̇2
(154)
a
Now let
so that
Z
b
(L0 − Q̇1 )η dt.
δΓ =
(155)
a
Setting δΓ = 0, the integrand must vanish, yielding L0 − Q̇1 = 0. Reversing though our
definitions of the Qn , this yields
...
L0 − L̇1 + L̈2 − L 3 = 0.
(156)
Written out fully in terms of the derivatives of the Lagrangian, this becomes
∂L
d ∂L
d2 ∂L
d3 ∂L
−
+ 2
− 3 ... = 0.
∂q
dt ∂ q̇
dt ∂ q̈
dt ∂ q
(157)
These results suggest that, for L = L(q, q̇, q̈, . . . , q (n) ), the generalized Euler-Lagrange equation would be
n
X
j=0
(−1)j
dj ∂L
= 0.
dtj ∂q (j)
28
(158)
This assertion may be proved by induction. We have established that
3
X
(j)
(−1)j Lj = 0
(159)
j=0
is the Euler-Lagrange equation for a Lagrangian that includes the third derivatives, q (3) , and
where for any quantity Λ,
Λ(n) ≡
dn Λ
.
dtn
(160)
We need to show that the generalized Euler-Lagrange equation for Lagrangians with Qn
necessarily applies when q (n+1) is included; in other word if
n
X
(−1)j L(j)
n = 0
(161)
(−1)j L(j)
n = 0
(162)
j=0
then
n+1
X
j=0
as well. Consider therefore the functional
Z b
L(q, q̇, q̈, . . . , q (n) , q (n+1) ) dt.
Γ≡
(163)
a
Taking the usual variation, and assuming the variation η vanishes on the boundaries, and
using the result of part (b) (where n ≤ 3), we obtain
!
Z b X
Z b
n
j (j)
δΓ =
η dt +
Ln+1 η (n+1) dt.
(−1) Lj
a
(164)
a
j=0
Our task is to use integrations by parts to reduce the η (n+1) term to something proportional
to η, so the result can join the other integral with the η factored out. Consider
d
(1)
Ln+1 η (n) = Ln+1 η (n) + Ln+1 η (n+1)
dt
(165)
so that
(1)
Ln+1 η (n+1) = −Ln+1 η (n) + Boundary T erm.
(166)
Second integration: to further reduce η (n) to η (n−1) , do another integration by parts. This
time consider
d (1) (n−1) (2)
Ln+1 η
= Ln+1 η (n−1) + L(1) η (n)
dt
29
(167)
which produces
(2)
Ln+1 η (n+1) = (−1)2 Ln+1 η (n−1) + Boundary T erms.
For a third integration by parts, examine
d (2) (n−1) (3)
(2)
Ln+1 η
= Ln+1 η (n−2) + Ln+1 η (n−1)
dt
(168)
(169)
which turns Ln+1 η (n+1) into
(3)
Ln+1 η (n+1) = (−1)3 Ln+1 η (n−2) + Boundary T erms.
(170)
Thus after k integrations by parts, we have
(k)
Ln+1 η (n+1) = (−1)k Ln+1 η [n−(k−1)] + Boundary T erms.
(171)
To make this result proportionial to η we set n − k + 1 = 0 (because η (0) = η), which requires
k = n + 1. Therefore we may write
Ln+1 η (n+1) = (−1)n+1 L(n+1) η + Boundary T erms.
(172)
When placed into the equation δΓ = 0, and with the vanishing of the boundary terms, Eq.
(164) becomes
Z
b
δΓ =
a
Z
=
a
n
X
!
(n+1)
n+1
(−1)n L(n)
Ln+1
n + (−1)
η dt
j=0
b
n+1
X
!
(j)
(−1)j Lj
η dt.
j=o
Therefore, setting δΓ = 0 gives the generalized Euler-Lagrange equation for a Lagrangian
that contains derivatives up to q (N ) :
N
X
(j)
(−1)j Lj = 0,
(173)
j=o
which says
N
∂L
d ∂L
d2 ∂L
∂L
N d
= 0.
−
+ 2
− . . . + (−1)
N
∂q
dt ∂ q̇
dt ∂ q̈
dt ∂q (n)
(174)
3.17 Here we consider a functional with a variable limit on the definite integral; for
definiteness let it be the upper endpoint, so that
Z T
Γ(T ) =
L(t, q µ , q̇ µ ) dt.
a
30
(175)
. What does δΓ = 0 give now? Recall we consider the transformation q 0µ = q µ + εη µ , leaving
t unchanged. In the derivation of the Euler-Lagrange equation, for fixed a and b we saw
that
Z b
δΓ =
a
∂L
d ∂L
−
µ
∂q
dt ∂ q̇ µ
∂L µ
η dt +
η
∂ q̇ µ
b
.
(176)
a
In the original derivation, η µ (b) = 0 and η µ (a) = 0. Here we are saying that, my replacing
fixed b with variable T, the upper boundary term survives:
Z b
∂L
d ∂L
∂L
η dt +
η µ (T ) = 0.
−
δΓ =
µ
µ
µ
∂q
dt
∂
q̇
∂
q̇
a
T
(177)
With no further assumptions, this is all we can say. But anticipating what is to come, if
both endpoints a and b are variable, and if we impose the Euler-Lagrange equation
∂L
d ∂L
−
= 0,
µ
∂q
dt ∂ q̇ µ
(178)
then we are rewarded with a conservation law:
pµ (b)η µ (b) = pµ (a)η µ (a),
(179)
i.e., pµ η µ = const. where pµ ≡ ∂L/∂ q̇ µ denotes the momentum canonically conjugate to q µ .
This result will be generalized later to gransformations of the form
t0 = t + ετ,
q 0µ = q µ + εζ µ
where τ = τ (t, q µ ) and ζ µ = ζ µ (t, q ν ). Then pη = const. will be seen as a special case of a
conservation law with far greater generality that involves ζ and τ in a superposition of the
canonical momentum and Hamiltonian:
pζ − Hτ = const.
(180)
This will be Noether’s theorem.
(Exercise 3.18) (a) In this problem we work out some equations for geometrical optics
that resemble the equations of Newtonian mechanics. Recall that Fermat’s principle says
31
that the functional for a light ray moving in the xy plane, through a medium of variable
refractive index n(x, y) is:
1
∆t =
c
s
b
Z
n(x, y) 1 +
a
dy
dx
2
dx.
(181)
Introduce an arbitrary parameter s (e.g., an arc length along the ray), and let primes denote
derivatives with respect to s. Now the functional may be written parametrically as
Z
1 b p 02
∆t =
n x + y 02 ds
c a
Z
1 b
dr
=
n(r)| | ds
c a
ds
where r denotes the position vector with components x = x1 and y = x2 .
R
(b) Defining the functional as c∆t = L dt, the Lagrangian is
L(x, y) = n(r)|
dr
| ≡ nr0 .
ds
(182)
The canonical momentum components are
pk =
∂L
dr0
nx0k
=
n
=
.
∂x0k
dx0k
r0
(183)
Notice this means that
x0k
r 0 pk
=
n
(184)
or, in terms of vectors, p = nr̂, and thus
p2 ≡ p · p = n2 .
(185)
(c) The Hamiltonian is
H ≡ pk x0k − L
0 k
rp
= pk
− nr0
n
2
p
0
= r
−n
n
r0 2
p − n2 = 0.
=
n
Curiously, the Hamiltonian is numerically equal to zero, even though it is a function of
canonical momenta and position coordinates. (A similar situation may occur for a particle
32
moving in a 1/r central potential. Such potentials give conic sections for the particle’s
trajectory, with the central potential’s source body located at a focal point of the conic. In
the case of the parabolic trajectory, E = 0 numerically even though the energy is a function
of velocity (in the kinetic energy) and radius (in the potential energy); see any intermediate
mechanics textbook.)
(d) If we impose H = K + U and L = K − U where K denotes a ”geometrical optics
kinetic energy” K ≡ 12 r02 (there being no mass term for geometrical optics), then U follows
at once. For instance, H − L = 2U so that
1
U = − nr0 .
2
(186)
But potential energy is supposed to be a function of the coordinates, not the “velocities.”
Since L = nr0 then L = K − U becomes
1
1
nr0 = r02 + nr0
2
2
(187)
1
U = − n2 (r).
2
(188)
which implies that n = r0 . Therefore
Alternatively, since the Hamiltonian is formally a function of momenta instead of velocities,
when imposing H = K + U it would have been better to write K = p2 /2. Here we see
immediately that p = r0 ; more about this point below. With H = p2 /2 + U and p2 = n2 , it
again follows that U = −n2 /2. Notice we have the three-way relation p2 = r02 = n2 . Writing
the Lagrangian as
1
1
L = r02 + n2
2
2
(189)
produces as the Euler-Lagrange equation
n
∂L
dx0
= k
∂xk
ds
(190)
d2 r
dx2
(191)
or in vector notation,
−∇U =
where U = −n2 /2. This resembles Newton’s second law, but applied here to geometrical
optics.
(e) Let the light ray travel in the xy plane, with a piecewise variable refractive index n(r) =
33
n1 = const. for y > 0 and n(r) = n2 = const. for y < 0. Then −∇U = r00 gives x00 = 0, or
dx0 /ds = 0, which integrates to dx/ds = const. across the y = 0 plane. Letting a light ray
encounter the y = 0 boundary with angles θ measured from the normal at the origin, and
noting that for length r along the ray above and below this plane that x = r sin θ, we have
dr
dr
sin θ1 =
sin θ2 .
(192)
ds 1
ds 2
But r0 = n, yielding Snell’s law,
n1 sin θ1 = n2 sin θ2
(193)
(f) In the xy plane, we are given n = no + αy. Now −∇(n2 /2) = r00 becomes, for the
x -component, x00 = 0. Two integrations, along with with the boundary condition s = 0
when x = 0, gives x = βs where β = const. Hence we may write s in terms of x, s = x/β.
The y-component gives, after some rearrangement,
y 00 − α2 y = no α.
(194)
00
yH
− α2 yH = 0,
(195)
yH = A cosh(αs) + B sinh(αs)
(196)
The homogeneous equation,
has the solution
where A and B are constants. From yH the solution to the homogeneous equation can be
constructed. Towards this end we allow A and B to become functions of s, and consider the
solution
y(s) = A(s) cosh(αs) + B(s) sinh(αs).
(197)
With the supplementary condition (see “variation of parameters” in any differential equation
textbook)
A0 cosh(αs) + B 0 sinh(αs) = 0
(198)
(where primes denote derivatives with respect to s), in the inhomogeneous differential equation for y the A can be eliminated, leaving the differential equation B 0 = no cosh αs, which
integrates to
B=
no
[sinh(αs) + γ]
α
34
(199)
where γ = const. With this, the supplementary condition becoems A0 = −no sinh(αs),
which integrates to
A=−
no
[cosh(αs) + δ]
α
(200)
where δ is another constant. Collecting these results, y becomes
y=
no
[γ sinh(αs) − δ cosh(αs) − 1].
α
(201)
Since y = 0 at x = 0 (and thus at s = 0), it follows that δ = −1. With the boundary
condition y 0 = 0 at x = 0 we find that γ = 0. Thus we have the equation of the ray:
no
αx
y=
cosh
−1 .
(202)
α
β
3.19 Here we write a generic Euler-Lagrange equation for a Newtonian particle moving in
a possibly time-dependent potential, in Euclidean space mapped by an arbitrary coordinate
system. The Lagrangian may be parameterized in terms of the coordinate system’s metric
tensor gjk = gkj , according to
1
L = mgjk q̇ j q̇ k − U (q i )
2
(203)
where the gjk may be functions of coordinates in three-dimensional space. The EulerLagrange equation,
∂L
d ∂L
=
n
∂q
dt ∂ q̇ n
(204)
becomes
1
d 1
j k
j
k
j k
−∂n U + m(∂n gjk )q̇ q̇ =
mgjk δ n q̇ + q̇ δ n
2
dt 2
d
mgnk q̇ k
=
dt
= m(∂j gnk )q̇ j q̇ k + mgnk q̈ k
where ∂j ≡ ∂/∂q j and we have used the chain rule on the metric tensor components. Upon
transposing a term this becomes
1
k
j k
j k
−∂n U = m gnk q̈ + (∂j gnk )q̇ q̇ − (∂n gkl )q̇ q̇ .
2
35
(205)
Multiply by g `n :
1
j k
`n
j
`n
∂j gnk − ∂n gjk q̇ q̇ .
−g ∂n U = m g gnk q̈ + g
2
`n
(206)
Recognizing that g `n gnk δ ` k , this becomes
1 `n
`
`
j k
−∂ U = m q̈ + g (2∂j gnk − ∂n gjk ) q̇ q̇
2
1 `n
`
j k
= m q̈ + g (∂j gnk + ∂k gnj − ∂n gjk ) q̇ q̇
2
where symmetry under index exchange (∂j gnk )q̇ j q̇ k = (∂k gnj )q̇ k q̇ j has been used. Define
1
Γ` jk ≡ g `n (∂j gnk + ∂k gnj − ∂n gjk )
2
(207)
−∂ ` U = m q̈ ` + Γ` jk q̇ j q̇ k .
(208)
to obtain at last
Exercise 3.20 In the problem we do, for a free particle moving relativistically in nonEuclidean spacetime, what we did for a Newtonian particle in Ex. 3.19. We start with the
functional
Z
b
Z
τ (b)
p
gµν uµ uν dτ = max.
dτ =
∆τ =
τ (a)
a
where uµ ≡ dxµ /dτ . Our Lagrangian is L = L(xµ , uµ ) =
(209)
√
gµν uµ uν . It will be noticed that,
numerically, L = 1 although the Lagrangian is a function of the generalized coordinates and
velocities. The Euler-Lagrange equation
∂L
d ∂L
=
ρ
∂x
dτ ∂uρ
(210)
µ
ν
1 ∂gµν µ ν
d 1
∂u ν
µ ∂u
u u =
gµν
u +u
.
2L ∂xρ
dτ 2L
∂uρ
∂uρ
(211)
becomes
Since ∂uµ /∂uρ = δ µ ρ , this becomes
1 ∂gµν µ ν
d 1
µ
u u =
gρµ u .
2L ∂xρ
dτ L
(212)
Invoking the numerical value L = 1, this gives
1
duµ
(∂ρ gµν )uµ uν = (∂σ gρµ )uσ uµ + gρµ
.
2
dτ
36
(213)
Manipulating the summed-out dummy indices, this may be rearranged into the form
d 2 xµ
1
(214)
gρµ 2 = ∂ρ gµν − (∂ν gρµ ) uµ uν .
dτ
2
Multiply by g λρ and recall that g λρ gρµ = δ λ µ . This produces
d2 xµ 1 λρ
+ g (2∂ν gρµ − ∂ρ gµν ) uµ uν .
2
dτ
2
(215)
Using the symmetry of the metric tensor, the 2(∂ν gρµ )uµ uν term can be written (∂ν gρµ +
∂µ gρν )uµ uν so that
d 2 xµ
+ Γλρ uµ uν = 0
dτ 2
(216)
1
Γλ µν ≡ g λρ (∂µ gνρ + ∂ν gµρ − ∂ρ gµν ) .
2
(217)
where
Exercise 3.21 This exersise takes us through some early minimization principles that
precede Hamilton’s principle. We begin with John Bernoulli’s 1717 principle of virtual
work. In a system of N particles in static equilibrium, consider virtual displacements drn of
unspecified direction for the nth particle; that is why they are called ‘virtual.” The directions
are taken care of by a constraint described below. If the virtual work done on all particles
vanishes, then
N
X
Fn · drn = 0,
(218)
n=1
Now for the relative directions of the virtual displacements: they are subject to the collective
constraint
N
X
Cn drn = 0
(219)
n=1
where the Cn are constants.
(a) Consider an Atwood’s machine, two masses mass 1 and mass 2, that are connected by a
string of negligible mass draped over a frictionless peg. Eq. (218) becomes
m1 g dy1 + m2 g dy2 = 0.
(220)
But the constraint of Eq. (219) gives dy1 = −dy2 , since as one mass falls the other rises the
same amount. Thus m1 g = m2 g, the condition for the two masses to be in equilibrium.
37
(b) If Fn = −∇Un then the principle of virtual work becomes
0=−
X
∇Un · drn = −
n
where U =
P
n
X
dUn = −dU
(221)
n
Un denotes the sum of all the potential energies. Thus dU = 0 means the
system will be in equilibrium when U is a maximum or a minimum.
(c) We now turn to d’Alembert’s principle, which extends the principle of virtual work
from statics to dynamics.
Consider Newton’s second law applied to the nth particle:
Fn = mn an , where the forces are partitioned into “imposed” forces (such as gravity, electric
forces, etc.) and “constraint” forces (such as normal forces, forces that connect masses with
inextendible strings, etc.). Distinguishing these two types of forces with superscripts, with
d’Alembert we partition the net force as
(c)
Fn = (F(i)
n + F )n .
(222)
Thus Newton’s second law applied to the nth particle may be written
(c)
F(i)
n − mn an = −Fn .
(223)
Dotting this by the particle’s virtual displacement then summing the result over all particles,
and noting that the forces of constraint do no work, we obtain
X
F(i)
−
m
a
· drn = 0
n
n
n
(224)
n
To illustrate, consider a modified Atwood machine, where a block of mass m1 sits on a
frictionless plane making angle θ1 above the horizontal, and another mass m2 sits opposite
it on another frictionless plane at angle θ2 (the two planes intersect like the peak of a
roof). Let the two masses be connected by a non-stretching string of negilgible mass, with
a frictionless peg carrying the string over the apex where the two planes meet. Eq. (224)
becomes, for the two-mass system,
(m1 g sin θ1 − m1 a1 ) dr1 + (m2 g sin θ2 − m2 a2 ) dr2 = 0.
(225)
But by Eq. (219), dr1 = −dr2 , and because the string does not stretch, a1 = a2 ≡ a, so
with some algebra we obtain
a=g
m1 sin θ1 − m2 sin θ2
m1 + m2
38
,
(226)
(i)
the same as by more conventional methods of elementary mechanics. Furthermore, if Fn =
−∇U , and using the definitions of acceleration and velocity, Eq. (224) may be developed as
follows:
0 =
=
=
=
dvn
· drn
Σn −∇Un · drn − mn
dt
Σn [−dUn − mn (vn · dvn )]
1
2
Σn −dUn − d(mn v )
2
−Σn dEn
where En denotes the sum of the αth particle’s kinetic and potential energies. Therefore,
when the “impressed” forces are derivable from a potential energy, d’Alembert’s principle is
equivalent to the requirement that the system’s energy be a minimum or a maximum.
(d) Turning to Gauss’s principle of least constraint (1829), imagine that with no constraints the nth particle would go from point A to point B in time ∆t, but that with
constraints it would go in the same time from A to another point C. Gauss’s principle of
least constraint asserts that
G≡
X
mn (rBC )2n = min.,
(227)
n
where
rBC ≡ rAC − rAB ,
(228)
the difference between the actual displacement and the displacement that would occur in the
absence of constraints. To illustrate, the principle, return to the simple Atwood machine, two
masses connected by an ideal string looped over a frictionless peg. Without the contraining
peg and string, both masses would fall freely when released from rest, so that, with a y-axis
pointed vertically downward, masses 1 and 2 would in time t drop the distance
1
yo1 = gt2 = yo2
2
(229)
where the sub-o denotes the absence of constraints. With the constraining peg and string,
the masses drop instead the distances
1
y1 = a1 t2
2
1 2
y 2 = a2 t .
2
39
But the non-stretching string connecting the two masses means that when one moves up, the
other moves down the same distance, so that a1 = −a2 ≡ a. The differences in displacements
with and without constraints are, for masses 1 and 2,
1
(a − g)t2
2
1
= (−a − g)t2 .
2
y1 − yo1 =
y2 − yo2
Now we can set up G:
1
G = t4 [m1 (a − g)2 + m2 (a + g)2 ].
4
(230)
Setting dG/da = 0 gives a familiar result,
a=g
m1 − m2
m1 + m2
.
(231)
Exercise 3.22 We seek a Lagrangian that will produce, in the Euler-Lagrange equation,
the fourth-order differential equation
W (x) = Y I
d4 y
dx4
(232)
where Y and I are constants, and W (x) is a given function.
(a) Recalling the approach of Exercise 3.16, we know that
d ∂L
d2 ∂L
d3 ∂L
∂L
−
+ 2
− 3 ... = 0.
∂y
dt ∂ ẋ dt ∂ ẍ dt ∂ x
(233)
Since our differential equation features a fourth derivative, the Lagrangian must contain a
...
third derivative. Therefore we seek a Lagrangian of the form L = L(x, ẋ, ẍ, x ).
(b) To find a Lagrangian for the cantilevered beam, the independent variable t in part (a)
becomes x, the dependent variable x(t) becomes y(x), and ẋ = dx/dt becomes y 0 = dy/dx.
The fourth-order Euler-Lagrange equation transcribed into these variables has the form
∂L
d ∂L
d2 ∂L
d3 ∂L
−
+
−
= 0.
∂y
dx ∂y 0 dt2 ∂y 00 dt3 ∂y 000
40
(234)
The first term will peel off the W (x) if the Lagrangian includes the term yW (x). The
derivative of y 000 and a third derivative of y 0 will produce fourth derivatives of y. Thus we
try the Lagrangian
L(x, y 0 , y 000 ) = yW (x) + Ay 0 y 000
(235)
where A is a parameter to be fit to the cantilevered beam equation. The Euler-Lagrange
equation Eq. (234) gives
d3
d
(Ay 000 ) − 3 (Ay 0 )
dx
dx
d4 y
= W (x) − 2A 4 ,
dx
0 = W (x) −
which yields the desired equation by setting A = 12 Y I.
41
WHEN FUNCTIONALS ARE INVARIANT
CHAPTER 4: Invariance
Exercise 4.1. A nonrelativistic particle of mass m and electric charge q moves through
an electromagnetic field described by the potentials V (t, r) and A(t, r). The Lagrangian is
1
L = mṙ2 + qv · A − qV.
2
(236)
(a) We are to determine whether the functional is invariant under, first, the simplest Galilean
transformation,
t0 = t
x 0 = x − vo t
y0 = y
z 0 = z;
(b) then, second, under the simplest Lorentz transformation,
t0 = γ(t − vx),
x0 = γ(x − vt)
y0 = y
z0 = z
where γ = (1 − v 2 )−1/2 wih v the relative velocity between frames (here using units where
c = 1). We can test for invariance by seeing if the following equations are satisfied: either
the definition of invariance,
L0
df
dt0
−L=ε
dt
dt
(237)
to first order in ε, or, equivalently, through the invariance identity,
∂L µ
∂L
ζ + pµ ζ̇ µ +
τ − H τ̇ = f˙.
µ
∂x
∂t
(238)
Let us begin with the definition of invariance. The functional will be invariant if and
only if, to first order in ε = vo ,
L0
dt0
df
− L = vo .
dt
dt
42
(239)
The left-hand side becomes
0 0
1
1
0 dt
02
0
0
0 dt
2
L
− L = mv + q(v · A ) − qV
− mv − q(v · A) + qV .
dt
2
dt
2
(240)
The given Galilean transformation gives for the transformation of the particle’s velocity,
v 0 = v − vo
(241)
where vo = îvo . The transformations of the potentials require a little more work, because
V 0 ≡ V (t0 , r0 ) and A0 ≡ A(t0 , r0 ). To get at these, perform a Taylor series expansion about
r0 = r (noting that t0 = t, so the t-dependence may be suppressed):
∂V
0
0µ
µ
V (r ) = V (r) + (x − x )
+ ...
∂x0µ o
∂V
= V − vo t
+ ...
∂x
= V − t(vo · ∇V ) + . . . .
Similarly, for A0 we obtain
A(r0 ) = A − t(vo · ∇)A + . . . .
(242)
Now the left-hand side of the invariance identity becomes
0
1
∂A
∂V
0 dt
2
L
− L = m(v − vo ) + q(v − vo ) · A − vo t
− q V − vo t
dt
2
∂x
∂x
1
− mv2 − qv · A + qV.
2
Several terms drop out, leaving
L0
dt0
− L = −vo î · [qt∇(v · A − V ) + mv + qA]
dt
= −vo î · u
where
u ≡ tq∇(v · A − V ) + mv + qA.
(243)
Let us unpack some terms in u a little more. Using the vector identity for the gradient of a
scalar product of vectors, one may write
∇(v · A) = v × (∇ × A) + A × (∇ × v) + (v · ∇)A + (A · ∇)v.
43
(244)
Because the Lagrangian formalism treats xµ and ẋµ as independent variables, derivatives of
velocity with respect to spatial coordinates vanish. Furthermore, the curl of A is a magnetic
field B and qv × B is the magnetic force F(m) on the charge q. This leaves
∇(v · A) = F(m) + (v · ∇)A.
(245)
In addition, taking the total time derivative of A(t, r) gives
∂A ∂A µ
dA
=
+ µ ẋ
dt
∂t
∂x
∂A
+ (v · ∇)A.
=
∂t
Now u may be written
(m)
u=t F
Noting that −∇V −
∂A
∂t
dA
∂A
+q
−q
− q∇V
dt
∂t
+ mv + qA.
(246)
is the electric field E and F (e) = qE is the electric force on charge
q, u becomes
u = t(F(e) + F(m) ) + mv + qA.
(247)
Now we see that the functional will be invariant under the Galilean transformation, in the
sense that L0 (dt0 /dt) − L = 0, if and only if the electric force, magnetic force, and canonical
momentum all have no x -component, which would be components parallel to the relative
motion between the two reference frames. In other words, for the functional to be invariant,
the electromagnetic interactions and particle motion must occur in directions perpendicular
to the relative motion. However, the functional can still be invariant, in the sense that
L0 (dt0 /dt) − L = vo f˙ if we identify f˙ = −î · u.
It is instructive to consider the invariance identity too. Let I denote the left-hand side of
the invariance identity. The functional is invariant under a given transformation if and only
if I = f˙. For both the Galilean and Lorentz transformations, vo plays the role of ε. In the
Galileo transformation, τ = 0, ζ 1 = −t, ζ 2 = 0, ζ 3 = 0. Inserting these into I, the nonzero
terms are
∂L
∂L
(−t) +
(−1)
∂x
∂ ẋ
∂
= −qt (v · A − V ) − (mẋ + qAx )
∂x
= −î · u
I =
44
where this u is the same vector we met previously.
Comparing the definition of invariance to the invariance identity, they give equivalent
results: to first order in ε = vo , they have produced
L
0
0 dt
dt
− L = −vo î · u,
I = −î · u.
Notice that −î · u is the derivative with respect to vo of the definition of invariance. This
differentiation was the crucial step in going from the definition of invariance to the invariance identity. To sum up, the functional for a charged particle moving nonrelativistically
through an electromagnetic field is invariant–in the context of the more liberal definition of
invariance–under the simplest Galileo transformation.
Turning now to the Lorentz transformation, we are to see whether or not I = 0, where
I≡
∂L
∂L
∂L µ
ζ + µ ζ̇ µ +
τ − H τ̇
µ
∂x
∂ ẋ
∂t
(248)
where H denotes the Hamiltonian,
H ≡ ẋµ
∂L
− L.
∂ ẋµ
(249)
The given Lorentz transformation is, in units where c = 1,
t0 = γo (t − vo x)
x0 = γo (x − vo t)
y0 = y
z0 = z
where vo denotes the velocity parallel to the x -axis of the primed frame relative to the
unprimed frame, and γo ≡ (1 − vo2 )−1/2 . Since vo plays the role of the parameter ε, to first
order in vo , γo ≈ 1. Therefore the generators are τ = −x, ζ 1 = −t, ζ 2 = 0, ζ 3 = 0. The
surviving terms of I are:
I = −t
∂L
∂x
∂L
−
−x
∂ ẋ
∂L
∂t
+ ẋH.
The terms in I are:
∂L
∂A ∂V
= q v·
−
,
∂x
∂x
∂x
∂L
= mẋγ + qAx
∂ ẋ
45
(250)
where γ ≡ (1 − v 2 )−1/2 (without the subscript) denotes γ for the particle’s velocity relative
to the unprimed frame;
∂L
∂A ∂V
= q v·
−
,
∂t
∂t
∂t
H = mγ + qϕ.
Putting these back into our expression for I, the mẋγ term cancels, and after some rearrangment we find
∂A ∂V
I = −q Ax − ẋV + t v ·
−
∂x
∂x
dF
≡
dt
∂A ∂V
+x v·
−
∂t
∂t
where the last line identifies dF/dt in the context of the more liberal definition of invariance.
Therefore we have found I = Ḟ for a complicated function Ḟ . Let us see if this is consistent
with the invariance definition. Towards this end we must see whether or not L(dt0 /dt) − L =
vo Ḟ with the same Ḟ .
Let us begin with dt0 /dt:
γo (dt − vo dx)
dt0
=
dt
dt
= γo (1 − vo vx ) ≈ 1 − β
where β ≡ vo · v. For L0 we must evaluate
L0 = −
m
+ q(v0 · A0 − V 0 )
γ0
(251)
where
0
dr0
dx dy 0 dz 0
v =
=
,
,
dt0
dt0 dt0 dt0
= (1 − β)−1 (v − vo ) ≈ (v − vo )(1 + β).
0
Thus
2
v0 = (v − vo )2 (1 + β)2 ≈ v 2 − 2βγ −2
(252)
and again, to first order in vo ,
1
= (1 − v 02 )1/2 ≈ γ −1 (1 + β).
γ0
46
(253)
Denoting V 0 ≡ V (t0 , r0 ) and A0 ≡ A(t0 , r0 ), a Taylor series expansion of V 0 and A0 about
r0 = r gives
∂V
+ ...
∂t
∂A
A0 = A − t(vo · ∇)A − (vo · r)
+ ....
∂t
V 0 = V − t(vo · ∇V ) − (vo · r)
Putting everything together, to first order in vo we find, after much algebra,
0
∂A ∂V
∂A ∂V
0 dt
L
− L = −vo q Ax − vx V + t v ·
−
+x v·
−
dt
∂x
∂x
∂t
∂t
(254)
which is equivalent to the result of the invariance identity with the same Ḟ .
Exercise 4.2. Consider the damped harmonic oscillator of constant mass m oscillating
with a time-dependent spring constant k(t) and a damping force proportional to the velocity
with constant damping coefficient b. The equation of motion
mẍ + bẋ + kx = 0
(255)
corresponds to the Lagrangian
1
1
2
2
L(t, x, ẋ) = mẋ − k(t)x ebt/m .
2
2
(256)
To determine whether the functional based on the time integral of this Lagrangian is invariant under the rescalings t0 = t(1 + ε) and x0 = x(1 + εκx), invoke the invariance identity,
I = 0, where
I=
∂L
∂L
∂L
ζ+
(ζ̇ − ẋτ̇ ) +
τ + Lτ̇ ,
∂x
∂ ẋ
∂t
(257)
and see what this requires of κ. Working out the various partial derivatives of L, and
denoting K = 21 mv 2 , U = 21 kx2 , and λ = b/m, we find, upon setting I = 0,
κ=
K + U − te−λt ∂L
∂t
.
2(K − U )
(258)
Without further assumptions, κ cannot be a constant, which contradicts the problem’s
hypothesis. Let’s try another approach, relevant for what we’ll see later when invoking the
invariance identity to find generators that lead to conservation laws. Postulating that the
47
invariance identity must hold regardless of whatever x(t) and ẋ(t) happen to be at any time,
I = 0 can be rearranged into terms proportional to ẋ and terms proportional to x :
!
k̇
= 0,
K(2κ − 1) − U 2κ + 1 + λt + t
k
(259)
suggesting a pair of equations that will result by setting separately to zero the coefficients
of K and U. From the first we find κ = (1 − λt)/2 which again is not constant, nor is it
consistent with setting to zero the coefficient of U. Another approach would be to try κ = 0
for arbitrary x(t) and ẋ. Now the invariance identity requires λt = 1, contradicting the
original statement of the problem that b and m are constants.
It seems that no constant κ exists under the proscribed circumstances. But we shall
revisit a version of this problem (see Ex. 5.1, in which k = const.), whose solution will refer
back to this exercise.
Exercise 4.3. We are to show that the two versions of the invariance identity,
∂L
∂L µ
ζ + pµ ζ̇ µ +
τ − H τ̇ = 0
µ
∂q
∂t
(260)
and
µ
µ
−(ζ − q̇ τ )
∂L
− ṗµ
∂q µ
=
d
(pµ ζ µ − Hτ ))
dt
(261)
are equivalent. It will be recalled that
pµ ≡
∂L
∂ q̇ µ
(262)
denotes the momentum conjugate to q µ , and
H ≡ pµ q̇ µ − L
(263)
denotes the Hamiltonian. In what follows the indices are suppressed. First, let us show that
Eq. (260) implies Eq. (261). From L = L(t, q, q̇) we can write the ∂L/∂t term as
∂L
∂L
= L̇ −
q̇ − pq̈
∂t
∂q
∂L
= −Ḣ + ṗq̇ + pq̈ −
q̇ − pq̈
∂q
∂L
= −Ḣ + ṗq̇ −
q̇.
∂q
48
Now Eq. (260) becomes, after some regrouping,
(ζ − q̇τ )
∂L
d(Hτ )
+ pζ̇ + ṗq̇τ −
= 0.
∂q
dt
Noting that pζ̇ = d(pζ)/dt − ṗζ, this may be arranged further into Eq. (261):
∂L
d
(ζ − q̇τ )
− ṗ + (pζ − Hτ ) = 0.
∂q
dt
(264)
(265)
To finish showing the two expressions for the invariance identity are equivalent, we must
go the reverse way, starting with Eq. (261) and showing that it can be re-arranged into Eq.
(260). In Eq. (261), distribute the factors on the left-hand side and employ the product rule
for derivatives on the right-hand side. The ζ ṗ term cancels, which leaves, after factoring out
all terms with coefficient τ ,
∂L
+τ
−ζ
∂q
∂L
− q̇ ṗ + Ḣ = −H τ̇ + pζ̇.
q̇
∂q
(266)
Using Ḣ = ṗq̇ + pq̈ − L̇, and writing L̇ with the chain rule where L = L(t, q, q̇), the term
with the τ coefficient reduces to τ ∂L/∂t. Now Eq. (261) has become
−ζ
∂L
∂L
− pζ̇ − τ
+ H τ̇ = 0,
∂q
∂t
(267)
which is the same as Eq. (260).
Exercise 4.4. We are to fill in the steps in the derivation of the invariance identity
for the inhomogeneous (liberal) definition of invariance. The inhomogeneous definition of
Rb
invariance states that the functional a L dt is invariant, by definition, if and only if
L0
dt0
dF
−L=ε
+ O(εs ), s > 1.
dt
dt
(268)
The corresponding invariance identity reads
∂L
∂L
∂L
∂L
dF
ζ+
ζ̇ +
τ − ẋ
− L τ̇ =
.
∂x
∂ ẋ
∂t
∂ ẋ
dt
(269)
The derivation of the invariance identity from the definition proceeds for the inhomogeneous
case with logic identical to that of the homogeneous definition. In particular, we take the
derivative of the definition with respect to ε, then set ε = 0. In differentiating we must
49
remember to use the chain rule on derivatives of L0 ≡ L(t0 , x0 , ẋ0 ). Therefore we must
evaluate
0
d dF
d
0 dt
s
L
−L =
ε
+ O(ε .
dε
dt
dε
dt
0
(270)
Since s > 1, the evaluation of the right-hand-side is trivial (one might suppose that the
h
i
0
0
2 F (t0 )
definition of invariance would require, on the right-hand side, dεd ε dFdt(t0 ) = dFdt(t0 ) + ε ddεdt
0 .
But that second term vanishes when we set ε = 0:
0
d
dF
0 dt
L
−L =
.
dε
dt
dt
0
(271)
The evaluation of the left-hand side proceeds identically to the homogeneous case; see Section
4.2 of the text.
Exercise 4.5. In the text it was shown that if the functional is invariant, then the
invariance identity holds. Reverse the argument, and show that if the identity holds, then
the functional is invariant. We begin with the invariance identity (the homogeneous case
will suffice, and indices are suppressed) arranged in this form:
∂L
∂L
∂L
ζ+
τ+
(ζ̇ − q̇ ṫ) + Lτ̇ = 0.
∂q
∂t
∂ q̇
(272)
We are to show this to be equivalent to
L0
dt0
− L = O(εs ), s > 1,
dt
which is equivalent, to within an irrelevant additive constant, to
0 0
d dt
dL
+L
=0
dε 0
dε dt 0
(273)
(274)
where L0 ≡ L(t0 , q 0 , dq(t0 )/dt0 ). The transformations being considered are, to first order in
ε, t0 = t + ετ and q 0 = q + εζ. Returning to the invariance identity, in it we need to write in
terms of the primed coordinates the Lagrangian, generators, and their derivatives. Let us
begin with τ and ζ, noting that τ = [dt0 /dε]0 and ζ = [dq 0 /dε]0 . Now the invariance identity
may be written
∂L dq 0
∂L dt0
∂L
d dt0
+
+
(ζ̇ − q̇ τ̇ ) + L
= 0.
∂q dε 0
∂t dε 0 ∂ q̇
dt dε 0
(275)
Turning to τ̇ , we see that
0 d dt0
d dt
τ̇ =
=
.
dt dε 0
dε dt 0
50
(276)
To handle the ζ̇ − q̇ τ̇ we recall in our earlier derivation of the invariance identity that
dq 0
q̇ + εζ̇
dq + εζ
=
.
=
0
dt
dt + ετ
1 + ετ̇
(277)
After performaing a binomial expansion on the denominator, this gives an expression for
ζ̇ − q̇ τ̇ in terms of primed coordinates:
d
ζ̇ − q̇ τ̇ =
dε
dq 0
dt0
.
(278)
0
Now the invariance identity takes the form
0 ∂L dq 0
∂L dt0
∂L d dq 0
d dt
+
+
= 0.
+L
0
∂q dε 0
∂t dε 0 ∂ q̇ dε dt
dε dt 0
0
This is the same as
"
0 #
∂L0 dq 0 ∂L0 dt0
d dq 0
∂L0
d
dt
+ 0
+ dq0
+ L0
=0
0
0
∂q dε
∂t dε ∂( dt0 ) dε dt
dε dt
0
or, by the chain rule and the product rule for derivatives,
0
d
0 dt
L
= 0.
dε
dt 0
(279)
(280)
(281)
To this we may add zero in terms of quantities whose derivatives with respect to ε vanish:
0
dO(εs )
d
0 dt
−L =
L
, s > 1.
(282)
dε
dt
dε
0
0
Thus the invariance identity can be reversed, and turned into the operation of taking the
derivative, with respect to ε, of the quantity
L0
dt0
− L = O(εs ), s > 0
dt
(283)
followed by setting ε = 0.
Exericse 4.6. In this problem we are to show from the invariance identity that central
force motion is invariant under a change of latitude and longitude coordinates. In elementary
language, since the force F = F(r) is central, the torque r × F vanishes, which means the
angular momentum ` is constant. Taking the direction of ` to define the z -axis, in spherical
coordinates the Lagrangian reads
1
1
L(r, ṙ, ϕ̇) = mṙ2 + mr2 ϕ̇2 − U (r)
2
2
51
(284)
where U denotes the potential energy. Clearly this Lagrangian is unchanged if ϕ and θ are
shifted by arbitrary constants. But in the spirit of exercising the Lagrangian formalism, for
a central potential we write
1
L = mv · v − U (r).
2
(285)
Our task is to see how the asserted indepenence of latitude and longitude emerges from the
invariance identity. The identity can be written
∂L
∂L µ ∂L
τ + µ (ζ̇ µ − q̇ µ τ̇ ) + Lτ̇ = 0.
ζ +
µ
∂q
∂t
∂ q̇
(286)
The transformations have the form
t0 = t + ετ
q 0µ = q µ + εζ µ .
In spherical coordinates, a particle’s velocity is given most generally by
v = ṙr̂ + rθ̇θ̂ + rϕ̇ sin θ ϕ̂.
(287)
From this we obtain a general expression, in spherical coordinates, for the particle’s angular
momentum:
` ≡ r × (mv) = mr2 θ̇ϕ̂ + mr2 sin2 θ (−θ̂)
(288)
`2 = m2 r4 (θ̇2 + ϕ̇2 sin2 θ).
(289)
and thus
Hence the Lagrangian may be written
1
`2
L = mṙ2 +
− U (r).
2
2mr2
(290)
In this formalism we have not yet demonstrated that ` is constant. We do notice, however,
that its `ϕ component is constant, because ∂L/∂ϕ = 0. Thus
const. =
∂L
2
= mr2 ϕ̇ sin2 θ = mr⊥
ϕ̇
∂ ϕ̇
(291)
where r⊥ ≡ r sin θ denotes the perpendicular distance from the z -axis to the particle’s
location. Since `ϕ = const., if θ changes then so does ϕ̇, so we can choose θ at our convenience
(which also follows from spherical symmetry). Let us choose θ = π/2. Then the Lagrangian
becomes the familiar one of Eq. (284).
52
Now consider changes of the latitude and longitude coordinates only:
t0 = t
r0 = r
π
θ0 =
+ εζ θ
2
ϕ0 = ϕ + εζ ϕ
so that τ = 0, ζ r = 0, and ζ θ and ζ ϕ are constants. The surviving terms on the left-hand
side of the invariance identity are
∂L θ ∂L ϕ
ζ +
ζ .
∂θ
∂ϕ
(292)
Both derivatives vanish, because L does not depend explicitly on either θ or ϕ. Thus
the invariance identity is satisfied, and the system invariant under a change of latitude or
longitude. We also confirm, consistent with the observation that r × F = 0, that
2 2
∂L
∂L
2
` =
+
= const.
∂ ϕ̇
∂ θ̇
(293)
Exercise 4.7 Show how a finite continuous transformation can be built incrementally
from a sequence of infinitesimal ones. To first order in ε, consider
t0 = t + ετ,
q 0µ = q µ + εζ µ .
For convenience, arrange these variables and generators into column vectors




t
τ




 q1 
 ζ1 








 . 
 . 




|Xi ≡ 
|Ki ≡ 
,
.
 . 
 . 








 . 
 . 




N
N
q
ζ
53
(294)
(a) Let |Ki ≡ M |Xi define a matrix M. For example, in the simplest Lorentz transformation,
to first order in the relative velocity ε,
t0 = t − εx
x0 = x − εt
y0 = y
z0 = z
so that |Ki = M |Xi in this particular instance becomes

 
 
−x
0 1 0 0
t

 
 
 −t   1 0 0 0   x 

 
 

=
 .
 0   0 0 0 0  y 

 
 
0
0 0 0 0
z
(295)
Consider now the infinitesimal transformation in matrix notation:
|X 0 i = (1 + εM )|Xi
= |Xi + εM |Xi
= |Xi + ε|Ki.
(b) Consider a second transformation on |X 0 i with the same generators:
|X 00 i = (1 + εM )|X 0 i
= (1 + εM )(1 + εM )|Xi
= (1 + εM )2 |Xi.
(c) Continuing this process with successive transformations, let |X (n) i denote n primes,
signifying the coordinate vector that results from n successive transformations. Thus
|X (n) i = (1 + εM )n |Xi.
Let ε = nθ , so that
|X
(n)
n
θM
i= 1+
|Xi.
n
In the limit as n → ∞, this becomes
|X (∞) i = lim (1 +
n→∞
θM
= e
|Xi.
54
θM n
) |Xi
ε
(296)
(297)
(d) Why does the definition of invariance in terms of infinitesimal transformations lose
nothing for finite continuous transformations? For an arbitrary transformation q 0 = Q(ε, t, q)
(suppressing indices), the transformation function Q can be expanded about ε = 0, provided
that Q is a differentiable function of ε where Q(0, t, q) = q. Thus
dQ
0
+ ...
q =q+ε
dε 0
(298)
which works for any continuously-parameterized transformation because the derivatives exist. It will not work for discrete transformations, such as time-reversals, space inversions,
charge conjugations, etc.
Exercise 4.8. Consider the Lagrangian for a Newtonian particle moving in one spatial
dimension x in the presence of a time-dependent potential:
1
L(t, x, ẋ) = mẋ2 − U (t, x).
2
(299)
Under the rescalings t0 = t + εt and x0 = x + 12 εx, so that τ = t and ζ = 12 x, what condition
on U (t, x) is necessary to make the functional invariant? We require the invariance identity
to hold:
∂L
∂L
∂L
ζ̇ +
ζ+
τ − H τ̇ = 0
∂x
∂ ẋ
∂t
(300)
where H denotes the Hamiltonian. For this Lagrangian, K = K(ẋ) and U = U (t, x). Noting
that H is numerically equal to 12 mẋ2 + U , with the given generators the kinetic energy term
drops out of the invariance identity, reducing it to
x ∂U
∂(tU )
+
= 0.
2 ∂x
∂t
(301)
This constraint must be satisfied if the functional is to be invariant with generators τ = t
and ζ = x/2. (See Section 5.4 on the inverse problem-finding generators that give invariance
of a given Lagrangian. See the similar situations of Exercises 5.5, 5.7, and 5.12.) If, as is
commonly the case, ∂U/∂t = 0, then Eq. (301) reduces to
x dU
+U =0
2 dx
(302)
which integrates to
U=
55
k
x2
(303)
where k = const. More generally, if U contains time dependence, we could try separation of
variables, and write U (t, x) = f (x)g(t). Now Eq. (301) becomes
ġ
x f0
+ t = −1
2f
g
(304)
where f 0 ≡ df /dx and ġ ≡ dg/dt. Introducing a separation constants α, this separates into
x f0
= α
2f
ġ
t = −(1 + α).
g
These integrate to U = kx2α t−(1+α) .
56
Chapter 5
Exercise 5.1 The Lagrangian
L=
1 2 −λt
1
2
mẋ − kx e
2
2
(305)
(where λ ≡ b/m) gives in the ELE the equation of motion for the damped oscillator,
mẍ − bẋ − kx = 0
(306)
where m, k, and b are constants. Within the context of the transformations t0 = t + ετ and
x0 = x + εζ, we seek generators τ and ζ that will make the functional invariant and thus
lead to Noether conservation laws. Towards this end we impose the invariance identity,
∂L
∂L
∂L
ζ+
(ζ̇ − ẋτ̇ ) +
τ + Lτ̇ = 0.
∂x
∂ ẋ
∂t
(307)
Noting that the identity must hold whatever the velocity at any time, we will write the
indentity as a polynomial in ẋ, then set to zero each coefficient of the various powers of ẋ.
Towards that end we must write τ̇ and ζ̇ using the chain rule,
τ̇ =
∂τ
∂τ
+
ẋ
∂t ∂x
(308)
and similarly for ζ̇. Evaluating the various derivatives of L that appear in the identitity,
after much algebra we obtain
0
1 ∂τ 2
1
x
= ẋ −kxζ − kλτ x − k x
2
2 ∂x
∂ζ 1
∂τ
+ ẋ1 m − kx2
∂t 2
∂x
1
∂ζ
∂τ 1 ∂τ
2
+ ẋ
mλτ + m
−m
m
2
∂x
∂t + 2 ∂t
1 ∂τ
+ ẋ3 − m .
2 ∂x
0
Setting to zero the separate coefficients of the various powers of ẋ, we obtain the Killing
equations: For the zeroth power,
1
∂τ
ζ + x λτ +
= 0;
2
∂t
57
(309)
from the ẋ coefficient we obtain
m
∂ζ 1 x ∂τ
− kx
= 0;
∂t 2
∂x
(310)
from the quadratic power of velocity we find
λτ −
∂ζ
∂τ
+2
= 0;
∂t
∂x
(311)
and from the cubic term it follows that
∂τ
= 0.
∂x
(312)
This equation from the cubic term requires in the first-order Killing equation that ∂ζ/∂t = 0
so that ζ = ζ(x). From these results, the zeroth-order equation becomes
dτ
−2ζ = x λτ +
dt
(313)
and the second-order equation becomes
−2
dζ
dτ
= λτ − .
dx
dt
(314)
Differentiating the first of these with respect to x and comparing it to the second requires
that dτ /dt = 0 so that τ = const. ≡ to . From the zeroth-order equation we now have
1
ζ = − xλto .
2
(315)
Noting that to will multiply ε in the transformation equations, we may without loss of
generality set to = 1 and present the generators, τ = 1 and ζ = −bx/2m, corresponding to
the transformations
t0 = t + ε
b
0
x = x 1−
ε
2m
which describes a diminishing rescaling of the particle’s position variable under a time translation, as fitting for a damped oscillator.
Let us revisit the equations of Exercise 4.2, where for the same system, but with a timedependent spring constant, we considered the transformation
t0 = t + εt
x0 = x + εκx
58
and the challenge was to see if a constant κ exists that makes the system invariant. The
Lagrangian was almost the same as in Ex. 5.1; the only difference being that k = k(t). The
answer was, in general, that no satisfactory κ exists. If we return to Ex. 4.2, set τ = 1 and
k̇ = 0, the invariance identity there collapses to (2κ + λ)(K − U ) = 0, where K = mẋ2 /2
and U = kx2 /2. This is guaranteed to be satisfied if κ = −λ/2, the same result as obtained
here in Ex. 5.1 via the Killing equations.
Exercise 5.2 Given L = L(t, ẋ) = t cot ẋ2 , let ẋ2 = u.
(a) The canonical momentum and Hamiltonian are:
p=
∂L
= −2tẋ csc2 u
∂ ẋ
H = pẋ − L = −t[2u csc2 u + cot u]
(316)
(317)
(b) Impose the invariance identity
∂L
∂L
ζ + pζ̇ +
τ − H τ̇ = 0,
∂x
∂t
(318)
Multiply this by sin2 u then use the identity sin u cos u = 12 sin(2u) to write the identity as
−2tẋ(ζ̇ − ẋτ̇ ) +
1
sin(2u)[τ + tτ̇ ].
2
(319)
Use the chain rule on τ = τ (t, x), ζ = ζ(t, x), and perform a Taylor series expansion on
sin(2u), and write the identity as a polynomial in ẋ. Setting to zero the coefficients of
various powers of ẋ yields the Killing equations...
Exercise 5.3 (a) One of the Emden equations,
ẍ +
2ẋ
+ x5
t
(320)
is the Euler-Lagrange equation
∂L
d ∂L
−
=0
(321)
∂x dt ∂ ẋ
of what Lagrangian? Clearly we must have a term proportional to x6 . With some trial and
error one constructs
L=t
2
ẋ2 x6
−
2
6
59
(322)
as the necessary Lagrangian.
(b) Under the transformation t0 = t + ετ (t, x), x0 = x + εζ(t, x), imposing the invariance identity as a polynomial in velocity, using the chain rule on the time derivatives of
the generators, and setting to zero the coefficients of various powers of ẋ gives the Killing
equations:
1 τx
x ∂τ
+
6 ∂t 3 t
∂ζ x6 ∂τ
−
ẋ1 :
∂t
6 ∂x
t
∂τ
∂ζ
ẋ2 : τ −
+t
2 ∂t
∂x
∂τ
ẋ3 :
∂x
ẋ0 :
ζ+
= 0;
= 0;
= 0;
= o.
The solutions are τ = −2Ct and ζ = Cx where C is a separation constant. Along the
way, as an intermediate step paramaterize τ as τ = atn + b. Consistency between the
various equations requires a = 1 and b = 0, leading to the stated result.
With these generators, the Noether conservation law, pζ − Hτ = const. becomes
1 6
2
2
t xẋ + tẋ + tx = const.
(323)
3
which can be confirmed by differentiating.
Exercise 5.4 Under a transformation from a Newtonian inertial reference frame to one
rotating with constant angular velocity ω relative to the inertial one, it is claimed that the
Lagrangian
1
L = m[v + (ω × r)]2 − U (t, r)
2
(324)
gives the equation of motion in a rotating reference frame:
−∇U − 2m(ω × v) − mω × (ω × v) = m
d2 r
.
dt2
(325)
In these expressions, r and v are respectively the instantaneous position and velocity of the
particle relative to the rotating system’s coordinates.
60
(a) To show this Lagrangian gives the stated equation of motion, we write the EulerLagrange equation (subscripts only for coordinates suffice because in Euclidian geometry no
distinction needs to be made between xµ and xµ ),
d ∂L
∂L
=
.
∂xµ
dt ∂ ẋµ
(326)
In terms of coordinates and their velocities, and using the Levi-Civita symbol εijk for the
cross products, and summing over repeated indices, the Lagrangian takes the form
1
1
L = mẋα ẋα − mεαβγ xα ωβ ẋγ + m(ω 2 xα xα − ωα ωβ xα xβ ) − U (t, xα ).
2
2
(327)
The left-hand side of the ELE becomes
∂L
∂U
= −m(ω × v)µ + mω 2 xµ − m(r · ω)ωµ −
∂xµ
∂xµ
(328)
and for the right-hand side,
d ∂L
= mẍµ + m(ω × v)µ .
dt ∂ ẋµ
(329)
Together these give the equation of motion in the rotating frame.
(b) Impose the invariance identity,
∂L µ ∂L
∂L
ζ +
τ + µ (ζ̇ µ − ẋµ τ̇ ) + Lτ̇ = 0
µ
∂q
∂t
∂ ẋ
(330)
while recalling that
τ̇ =
∂τ
∂τ
+ ν ẋν
∂t ∂x
(331)
∂ζ µ ∂ζ µ ν
+ ν ẋ .
∂t
∂x
(332)
and
ζ̇ µ =
Gather all the powers of q̇ µ into a polynomial. For this particular Lagrangian we obtain a
structure of the form (supressing coordinate subscripts):
ẋ0 A + ẋ1 B + ẋ2 C + ẋ3 D = 0
(333)
which must hold whatever the velocity, so the coefficients A, B, C, and D must be set
separately to zero. Setting A = 0 gives, after some algebra,
∂(U τ )
1 ∂τ
∂ζ
2
−
− ζ · ∇U + m (ω × r) + m(ω × r) ·
+ (ω × ζ) = 0.
∂t
2 ∂t
∂t
61
(334)
Denoting ∂ν ≡ ∂/∂xν , setting B = 0 produces
∂ζν
1
2 2
2
+ m [(ω × ζ)ν + (ω × r)µ (∂ν ζ µ )] = 0.
∂ν m ω r − (ω · r) − U + m
2
∂t
(335)
Setting C = 0 yields
∂ν ζµ −
1 ∂τ
δµν = 0,
2 ∂t
(336)
For instances where µ 6= ν this shows that ∂ν ζ µ = 0, so that ζµ = ζµ (xµ , t) and therefore
Eq. (336) reduces to
1
∂µ ζ µ = τ̇
2
(337)
∂ν τ = 0
(338)
(no sum over µ). For D = 0 we have
which requires that τ = τ (t). With these results in hand, Eq. (335) becomes
1
∂ζ
+ (ω × ζ) + τ̇ (ω × r) = 0.
∂t
2
If we try the ansatz ζ µ = g µ (xµ )f (t), then the Eq. (337) can be separated, to give
1
ζ=
Cr + ro f (t)
2
(339)
(340)
and
Z
τ =C
f (t) dt
(341)
where C is a separation constant and ro an integration constant. Placing this into Eq. (339)
turns the latter into
f˙ 1
Cr + ro + C(ω × r) + ω × ro = 0,
f 2
(342)
Multiple solutions exist; we can try various possibilities. For instance if f (t) = const. = 1
then
1
Cr + r0
2
τ = Ct + to
ζ =
where to is another integration constant. On the other hand, if f (t) = αtn where α and n
are constants, then τ = αCtn+1 + tn+1
and ζ gets a time-dependent scaling. One interesting
o
possibility occurs by the ansatz ζ = Aω where A = const.. This leads to τ = const. and
thus the transformation r0 = r+εAω under the time translation t0 = t+ε. Other special-case
solutions may also be constructed.
62
(c) Let us construct the Noether conservation law for a solution to the Killing equations
found above. In the instance ζ =
C
r
2
+ ro and τ = Ct + to , the conservation law
p · ζ − Hτ = 0
(343)
C
p · r + p · ro − H(Ct + to ) = 0.
2
(344)
becomes
We can identify several cases. First, if ro = 0, C = 0 but to 6= 0 the situation describes a
time translation, and the Hamiltonian is conserved. Second, if ro 6= 0, C = 0, to = 0 then a
space translation results, with the component of momentum parallel to ro being conserved,
i.e., p · ro = 0. Third, if ro = 0, to = 0, C 6= 0, then
1
p · r − Ht = const.,
2
(345)
a statement that, in slowly-varying periodic systems, leads to so-called “adiabatic invariance”
(see Sec. 5.5 of the text).
Exercise 5.5 The Lagrangian is
1
L(t, q µ , q̇ µ ) = mgµν q̇ µ q̇ ν − U (t, q ν ).
2
(346)
With the transformation
t0 = t + ετ
q 0µ = q µ + εζ µ .
To impose invariance, we require the invariance identity to be satisfied:
I≡
∂L
∂L ρ ∂L ρ
ρ
ζ
+
ζ̇
−
q̇
τ̇
+
τ + Lτ̇ = 0.
∂q ρ
∂ q̇ ρ
∂t
(347)
Evaluating the various derivatives, including the chain rule for time derivatives of the generators, and gathering everything up into a polynomial in velocities, the invariance identity
becomes, in this instance,
∂U
∂τ
∂τ
∂ζ ν
µ
µ
0 = − (∂µ U )ζ +
τ +U
+ q̇ U µ + mgµν
∂t
∂t
∂q
∂t
ρ
∂ζ
1
1
∂τ
1
∂τ
µ ν
ρ
µ ν ρ
+ q̇ q̇ mgµρ ν + m(∂ρ gµν )ζ − mgµν
− q̇ q̇ q̇
mgµν ρ .
∂q
2
2
∂t
2
∂q
63
Setting to zero the coefficients of each power of velocity gives the Killing equations:
∂τ
= 0,
∂q ρ
gµρ
∂ζ ρ 1
∂τ
1
+ (∂ρ gµν )ζ ρ = 0,
− gµν
ν
∂q
2
∂t 2
∂τ
∂ζ ν
U µ + mgµν
= 0,
∂q
∂t
∂(U τ )
(∂µ U )ζ µ +
= 0.
∂t
(b) Apply these Killing equations to a particle moving in a central potential U = U (r)
in spherical coordinates (q 1 , q 2 , q 3 ) = (r, θ, ϕ). The first Killing equation requires τ = τ (t),
makes ∂τ /∂t = dτ /dt in the second and fourth equations, and reduces the third equation to
∂ζ ν
=0
∂t
(348)
so that ζ ν is a function of at most the spatial coordinates. Returning to the second equation,
for µ = ν = r it becomes
∂ζ r
1 dτ
=
;
∂r
2 dt
(349)
1 dτ
ζ r ∂ζ θ
+
=
;
r
∂θ
2 dt
(350)
∂ζ ϕ ζ r
1 dτ
+
+ ζ θ cot θ =
.
∂ϕ
r
2 dt
(351)
1 dτ
= C = const.
2 dt
(352)
for µ = ν = θ it says
and for µ = ν = ϕ
These imply that
so that τ = 2Ct + to where to is an integration constant. Now the equations for the spatial
generators become
∂ζ r
=C
∂r
(353)
so that ζ r = Cr + f (θ, ϕ) where f is an arbitrary function; next we have
∂ζ θ ζ r
+
= C.
∂θ
r
(354)
If we choose f (θ, ϕ) = 0 then ζ θ = θo = const. Finally for the spatial generators we have
∂ζ ϕ ζ r
+
+ ζ θ cot θ = C.
∂ϕ
r
64
(355)
Since central force motion occurs in a plane, let that be the θ = π/2 plane, from which it
follows that ζ ϕ = ϕo = const. Now the generators describe at time translation with rescaling,
a radial rescaling, and rotations:
τ = 2Ct + to ,
ζ r = Cr,
ζ θ = θo ,
ζ ϕ = ϕo .
If one substitutes these generators back into the invariance identity, it reduces to the fourth
Killing equation, which puts a constraint on the potential in order for the invariance to go
through.
(c) Now we turn to the conservation laws that follow from the generators of part (b). If
the functional is both extremal and invariant, then Noether’s theorem gives
pµ ζ µ − Hτ = 0
(356)
where the pµ are the canonical momenta and H the Hamiltonian. For instance, if the only
nonzero contribution to the generators is τ = to then H = const., and for central motion
the Hamiltonian is the mechanical energy, kinetic plus potential. Thus invariance under a
time translation gives conservation of energy. If ϕo is the only nonzero contribution to the
generators, then pϕ = mr2 ϕ̇ = const., and invariance under a rotation about the z -axis
produces conservation of angular momentum. If C is the only nonzero contribution to the
generators, then
1
mrṙ − Ht = const.
2
(357)
which sets up “adiabatic invariance” as discussed elsewhere in Chapter 5.
Exercise 5.6 A central force in rectuangular coordinates has the Lagrangian
1
L = m(ẋ2 + ẏ 2 + ż 2 ) − U (r)
2
65
(358)
where r =
p
x2 + y 2 + z 2 . Under a rotation through angle ε about the z -axis,
t0 = t
x0 = x cos ε + y sin ε ≈ x + εy
y 0 = −x sin ε + y cos ε ≈ y − εx
z0 = z
so the only nonzero generators are ζ x = y and ζ y = −x. To check for invariance we see if
the invariance identity,
∂L µ
∂L
∂L
τ − H τ̇ = 0
ζ + µ ζ̇ µ +
µ
∂x
∂ ẋ
∂t
(359)
is satisfied. The nonzero terms that enter the identity gives on the left-hand side of the
identity
−
∂U
∂U
y+
x = (r × F)z
∂x
∂y
(360)
where F = −∇U . Thus the invariance identity’s left-hand side becomes the z -component of
the torque exerted by a central force, which vanishes. With the invariance identity satisfied,
the Noether conservation law, pµ ζ µ − Hτ = const. gives
∂L x ∂L y
ζ +
ζ = mẋy − mẏx = Lz = const.,
∂ ẋ
∂ ẏ
(361)
the conservation of angular momentum about the z axis.
Exercise 5.7 Eq. (K0) in the text is part of Eqs. (K3)-(K0) of p. 96, which apply to
the Lagrangian
1
L = mẋ2 − U (t, x).
2
(362)
For this Lagrangian, p = mẋ, and H = p2 /2m + U which numerically equals 21 mẋ2 + U =
K + U . The transformation considered is
τ = Ct + to
1
ζ = Cx + xo .
2
If pζ − Ht is to be constant, then the Euler-Lagrange equations
−
∂L
= ṗ
∂ ẋ
(363)
∂L
= Ḣ.
∂t
(364)
66
have already been invoked. Writing out
d
(pζ − Hτ ) = 0
dt
(365)
ṗζ + pζ̇ − Ḣτ − H τ̇ = 0.
(366)
gives
Using the Euler-Lagrange equations, this becomes
∂L
C ẋ
∂L
1
2
ζ + mẋ
+
τ−
mẋ + U C = 0.
∂x
2
∂t
2
(367)
A kinetic energy term cancels, leaving
−
∂U
∂τ
∂U
ζ−
−U
= 0,
∂x
∂t
∂t
(368)
or
∂U
∂(U τ )
ζ+
= 0,
∂x
∂t
(369)
which is Eq. (K0).
Exercise 5.8 Show that
I
pdx = 2hKiT
(370)
for a free particle of momentum |p| = po and kinetic energy K in a box of width w : For the
periodic integral,
I
Z
pdx =
w
Z
0
po dx +
0
(−po )dx = 2po w.
(371)
w
Let T denote the period for one cycle. Then
po
2w
=
m
T
(372)
p2o
2m
(373)
or w = po T /2m, and thus
I
pdx = 2
T = 2KT.
Since po = const,, the kinetic energy equals its average and thus
I
pdx = 2hKiT.
(374)
H
pdx = 2po w changes at the
If the walls move slowly apart then the action integral A ≡
rate Ȧ = 2pẇ so that
Ȧ
ẇ
= .
A
w
67
(375)
Thus if ẇ/w << 1 then Ȧ/A << 1. More formally, the Lagrangian for the particle in the
box is L = 21 mẋ2 so the results of Exercise 5.7 hold, with U = 0. Thus the generators are
τ = Ct + to
1
ζ = Cx + xo
2
and (without the additive integration constants) the invariance identity becomes
xẋ − ẋ2 t = const.
Multiply by m/2 and average over one period:
Z
Z
m T
dt m T 2 dt
xẋ −
ẋ t = const.
2 0
T
2 0
T
The first term integrates to zero:
Z
m 2
m T dx dt
x
=
x (T ) − x2 (0) = 0
2 0
dt T
4T
(376)
(377)
(378)
since x(T ) = x(0). Since the total energy E is purely kinetic and is constant for elastic
collisions with the wall, the second term may be written
Z T
1
dt
−E
t = − ET = const.
T
2
0
(379)
Furthermore, since H = pẋ − L, integrating the Hamiltonian over one cycle gives
I
I
I
Hdt =
pẋdt − Ldt
I
I
I
E dt =
pdx − Ldt.
Because the integral of Ldt is invariant and we have shown that ET is adiabatically invariant,
H
it follows that pdx is adiabatically invariant.
Exercise 5.9 According to Old Quantum Theory, the quantized energies follow from
quantizing the action according to
I
pdx = nh
(380)
where the momentum p is a function of energy E, h is Planck’s constant, and n = 0, 1, 2, . . ..
For the simple harmonic oscillator of mass m oscillating with angular frequency ω,
1
p2
1
E = mω 2 A2 =
+ mω 2 x2
2
2m 2
68
(381)
so that
√
p = ±mω A2 − x2 .
Thus
I
A
Z
pdx = 4mω
√
A2 − xx dx = nh
(382)
(383)
0
which integrates to
mωA2 π = nh
(384)
E = En = nh̄ω
(385)
or
where h̄ = h/2π. Old Quantum Theory misses the “zero point energy” h̄ω/2 which emerges
in the eigenvalues of the Schrödinger equation for the simple harmonic oscillator, which are
En = (n + 21 )h̄ω.
Exercise 5.10 This exercise refers back to Example 5 (p. 87) in Ch. 5 , with
1
L = m(ẋ2 + ẏ 2 ) − mgy
2
(386)
where the conservation law with the more liberal definition of invariance,
pµ ζ µ − Hτ − F = const.
(387)
is examined under the Galilean transformation:
t0 = t
x0 = x − v o t
y0 = y
so the generators are τ = 0, ζ x = −t, ζ y = 0, and F = −mx + const. The conservation law,
px ζ x + py ζ y − Hτ − F = const.
(388)
−mẋt + mx = const.
(389)
mẋ = m(xt + xo ),
(390)
becomes
which can be rearranged into
69
in which we recognize the familiar expression x = xo + vt which applies to particles moving
with constant velocity and therefore constant linear momentum.
Exercise 5.11 We are to solve the Killing equation (K0),
∂U
∂(U τ )
+ζ
=0
∂t
∂x
(391)
dτ
ζ dU
+
=0
dt U dx
(392)
for the following scenarios:
(a) U = U (x): Eq. (K0) becomes
which may be solved by separation of variables, to yield
τ = Ct + to
ζ dU
= −C
U dx
where C is a separation constant and to an integration constant. If U (x) is given then ζ
follows at once because
ζ = −C
U
U0
(393)
where U 0 = dU/dx.
(b) If U = U (t) then Eq. (K0) becomes
U̇ τ + U τ̇ = 0
(394)
or
dU
dτ
=
U
τ
which integrates to τ = C/U where C = const.
−
(395)
(c) If U = U (t, x) = g(t) + f (x) then Eq. (K0) becomes
ġτ + (g + f )τ̇ + ζU 0 = 0
(396)
and thus
dτ
df
d(gτ )
+f
+ζ
=0
(397)
dt
dt
dx
which is not separable in general. However, it separates if dτ /dt = C = const. so that
τ = Ct + to . Then
ζ=
1
[ġ(Ct + to ) + gC − Cf ] .
f0
70
(398)
(d) Now suppose U (t, x) = g(t)f (x). Eq. (K0) becomes
f0
1
(ġτ + g τ̇ ) + ζ = 0
g
f
(399)
which is amendable to separation of variables, so that
ġ
τ + τ̇ = C
g
ζ = −C
f
.
f0
The time-dependent equation integrates to
C
τ=
g(t)
Z
t
g(t0 )dt0 .
(400)
0
(e) For the simple harmonic oscillator (allowing the possibility of a time-dependent spring
constant), U (t, x) = 12 k(t)x2 . Eq. (K0) requires
k̇
2ζ
τ + τ̇ +
= 0.
k
x
(401)
For the generic case with this Lagrangian we found
τ = Ct + to
1
ζ = Cx + xo
2
where C, to , and xo are constants. Now Eq. (K0) becomes
C + xxo
1 dk
= −2
.
k dt
Ct + to
(402)
Letting V = C + (xo /x), after integration this yields
k = ko (Ct + to )−2V
(403)
where ko = const. To have ∂k/∂x = 0 requires xo = 0. Therefore
k(t) =
ko
.
Ct + to
(404)
Exercise 5.12 A particle of mass m and electric charge q moves through an electromagnetic field with potentials V and A. Its Lagrangian is
r
v2
L = −mc2 1 − 2 + qv · A − qV.
c
71
(405)
For v << c this becomes
1
L = mẋµ ẋµ + q ẋµ Aµ − qV
2
(406)
(sum over repeated indices; no distinction between upper and lower indices is necessary in
Euclidean space).
(a) The components of the canonical momenta are
pµ =
∂L
= mẋµ + qAµ .
∂ ẋµ
(407)
The Hamiltonian, H = pµ ẋµ − L, is numerically equal to
1
H = mẋµ ẋµ + q ẋµ Aµ + qV.
2
(408)
(b) Under the transformation t0 = t+ετ and x0µ = xµ +εζµ , impose the invariance identity
and find the generators. Allowing for the time and spatial dependence of the generators,
the invariance identity,
∂L
∂L
− H τ̇ = 0,
ζµ + pµ ζ̇µ +
∂xµ
∂t
(409)
becomes
∂ζβ
∂τ
∂τ
∂τ
∂τ
∂ζβ
1
ẋα +
−
ẋα ẋβ −
ẋβ +
mẋβ ẋβ + q ẋβ Aβ − qV
ẋα +
(mẋβ + qAβ )
∂xα
∂t
∂xα
∂t
2
∂xα
∂t
∂V
∂V
∂Aα
∂Aβ
−q
−q
+τ q ẋα
+ ζα q ẋβ
= 0.
∂t
∂t
∂xα
∂xα
We arrange this into a polynomial in powers of velocity:
1 ∂τ
1 ∂τ
∂ζβ
0
=
ẋα ẋβ ẋβ − m
− m δαβ
+ ẋα ẋβ m
2 ∂xα
∂xα 2 ∂t
∂ζβ
∂τ
∂ζα
∂Aβ
+ ẋβ m
− qV
+q
Aα + ζα
∂t
∂xβ
∂xβ
∂xα
∂ζβ
∂(V τ
∂V
+ qAβ
−q
− qζα
.
∂t
∂t
∂xα
Setting to zero the coefficients of each power of velocity gives the Killing equations, listed
72
here in decreasing order from cubic on down:
ẋ3 :
∂ζβ
∂τ
δαβ − 2
= 0,
∂t
∂xα
ẋ2 :
1
ẋ :
∂τ
∂ζβ
− qV
+q
m
∂t
∂xβ
ẋ0 :
∂ζα
∂Aβ
Aα + ζα
∂xβ
∂xα
Aβ
∂τ
= 0,
∂xα
+ qτ
∂Aβ
= 0,
∂t
∂ζβ ∂(V τ )
∂V
−
− ζα
= 0.
∂t
∂t
∂xα
The ẋ3 Killing equation gives τ = τ (t), which in the ẋ2 equation implies
dτ
∂ζβ
δαβ = 2
.
dt
∂xα
(410)
For α 6= β this gives ζβ = ζβ (xβ , t). For α = β there results τ = Ct+to and ζβ = 21 Cxβ +g(t)
where C is a sepration constant, to an integration constant and g(t) an arbitrary function
of time. In the simplest case g(t) = 0, the with the aid of the ẋ2 equation the ẋ1 Killing
equation becomes
1
1 ∂Aβ
∂Aβ
0 = δαβ Aα + xα
+t
2
2 ∂xα
∂t
(411)
which is the β th component of
0 = A + (r · ∇)A + 2t
∂A
.
∂t
(412)
Should the magnetic field be uniform, then A = 12 B × r and consequently (r · ∇)A = A,
whence the ẋ1 Killing equation becomes
∂(At)
=0
∂t
(413)
∂(V t) 1
+ r · ∇V.
∂t
2
(414)
and the ẋ0 equation becomes
When the functional is both invariant and extremal, Noether’s theorem gives the conservation law
1
p · r − Ht = const.
2
73
(415)
where p = mV + qA is the canonical momentum. In terms of the velocity, for a nonrelativistic particle the Noether conserved quanity is
1
1
1
− mv 2 t + mv · r + qA · r = const.
2
2
2
(416)
Now investigate adiabatic invariance by allowing the magnetic field to be spatially uniform
but vary slowly in time. With B = Bẑ in cylindrical coordinates (ρ, θ, z), the particle’s
position vector is r = ρρ̂ + zẑ so that in the ensuing helical motion, v = żẑ + ρθ̇θ̂. Let T be
the period for one orbit around the helix. Average the Noether conservation law over one
period, from t = 0 to t = T . The first term is
T
Z
1
1
−h mv 2 ti = −
2
T
0
1 2
mv t dt.
2
(417)
Assuming the magnetic field varies slowly over one helical orbit, so that the Faraday-induced
electric field E = −∂A/∂t ≈ 0, the kinetic energy varies little from its average value hKi
over one orbit so that
1
hKi
−h mv 2 ti ≈ −
2
T
Z
0
T
1
t dt = − hKiT.
2
(418)
The average of the v · r term is
m
1
mhv · ri =
2
2T
Z
T
0
m
v · r dt =
2T
Z
T
z ż dt =
0
m 2
z (T )
4T
(419)
where z(0) = 0. Since hKi varies little over one cycle, the velocities also vary little over one
cycle, so we may write z(T ) ≈ żT and thus
1
1
mhv · ri = mhż 2 iT.
2
2
(420)
The last term in the Noether conservation law vanishes because B is spatially uniform, for
which A = 12 B × r and therefore A · r = 0. In view of these results, the Noether conservation
law reduces to
T m 2 −
hKi − hż i const.
2
2
1
2
2 2
But hKi = 2 mhż + ρ θ̇ i which simplifies the conservation law to
−
mT 2 2
hρ θ̇ i = const.
4
(421)
(422)
Recall the Larmor frequency for a particle orbiting in a magnetic field, 2π/T = eB/m, which
allows us to swap T for B in the conservation law, giving one of the adiabatic invariants,
m hρ2 θ̇2 i
= const.
2 B
74
(423)
Since θ̇ = 2π/T = eB/m this can also be written in terms of the magnetic flux through the
helical trajectory,
πρ2 B = const.,
(424)
another adiabatic invariant of plasma physics. Finally, from the elementary definition of
magnetic dipole moment µ, as the product of circulating current and area, µ = (q/T )πρ2 ,
the previous result gives µ = const., the third adiabatic invariant of plasma physics.
(g) Here we confirm that the claimed Noether conserved quantity is indeed a constant by
showing that Ġ = 0 where
1
G ≡ p · r − Ht.
2
(425)
1
1
Ġ = ṗα xα + pα ẋα − Ḣt − H.
2
2
(426)
We must examine
Recalling the canonical momentum
p = mv + qA
(427)
(p − qA)2
1
H = pα ẋα − L =
+ qV = mv · v + qV,
2m
2
(428)
and Hamiltonian
since the Noether conservation law requires the Euler-Lagrange equations to hold, we may
use them to evaluate some of the terms, so that
ṗ = ∇L = q [∇(v · A) − ∇V ]
(429)
∂L
∂A ∂V
−Ḣ =
=q v·
−
.
∂t
∂t
∂t
(430)
and
With these results, we reassemble Ġ. Two kinetic energy terms cancel, leaving
Ġ
1
1
∂A
1
∂(V t)
= [∇(v · A)] · r + (v · A) + v ·
− (r · ∇V ) −
.
q
2
2
∂t
2
∂t
(431)
Noting the identity [∇(v · A)] · r = A · v, it follows that Ġ/q becomes
Ġ
=
q
∂(At)
∂t
1
∂(V t)
·v−
(r · ∇V ) +
.
2
∂t
75
(432)
By the Killing equations from the coefficients of the ẋ0 and ẋ1 terms in the invariance
identity, both terms vanish, and thus Ġ = 0.
Exercise 5.13 The Lagrangian of the damped oscillator,
1 2 −bt/m
1
2
mẋ − kx e
L=
2
2
(433)
gives by the Euler-Lagrange equation the equation of motion
dẋ
b
k
+ ẋ + x = 0.
dt
m
m
Multiply this by dx and integrate from x1 to x2 , which gives
Z x2
1
1 2
2
ẋ dx.
∆
mẋ + kx = −b
2
2
x1
(434)
(435)
The left-hand side is the change in mechanical energy E and the right side is the work done
by the resistive force −bẋ, an illustration of the work-energy theorem. The work done by
R
friction, Wf = bẋ dx, typically raises the temperature of the surfaces involved. If this work
is represented by a change in “internal energy U, then by including it energy is conserved:
∆(E + U ) = 0.
76